chapter solutions key 11 circles - shakopee.k12.mn.us · by thm. 11-1-3, rs = rt __x 4 = nx-6.3 x =...

Solutions KeyCircles11

CHAPTER

ARE YOU READY? PAGE 743

1. C 2. E

3. B 4. A

5. total # of students = 192 + 208 + 216 + 184 = 800

( 192 ____ 800

) · 100% = 24%

6. 216 ____ 800

· 100% = 27%

7. 208 + 216 _________ 800

· 100% = 53%

8. 11%(400,000) = 44,000

9. 27%(400,000) = 108,000

10. 19% + 13% = 32%

11. 32%(400,000) = 128,000

12. 11y - 8 = 8y + 1 3y - 8 = 1 3y = 9 y = 3

13. 12x + 32 = 10 + x11x + 32 = 10 11x = -22 x = -1

14. z + 30 = 10z - 15 30 = 9z - 15 45 = 9z z = 5

15. 4y + 18 = 10y + 15 18 = 6y + 15 3 = 6y y = 1 __

2

16. -2x - 16 = x + 6 -16 = 3x + 6 -22 = 3x x = - 22 ___

3

17. -2x - 11 = -3x - 1 x - 11 = -1 x = 10

18. 17 = x 2 - 3249 = x 2 x = ±7

19. 2 + y 2 = 18 y 2 = 16 y = ±4

20. 4 x 2 + 12 = 7 x 2 12 = 3 x 2 4 = x 2 x = ±2

21. 188 - 6 x 2 = 38 -6 x 2 = -150 x 2 = 25 x = ±5

11-1 LINES THAT INTERSECT CIRCLES, PAGES 746–754

CHECK IT OUT! PAGES 747–750

1. chords: −− QR ,

−− ST ; tangent: � �� UV ; radii: −− PQ ,

−− PS , −− PT ;

secant: � �� ST ; diameter: −− ST

2. radius of circle C: 3 - 2 = 1radius of circle D: 5 - 2 = 3point of tangency: (2, -1)equation of tangent line: y = -1

3. 1 Understand the ProblemThe answer will be the length of an imaginary segment from the summit of Mt. Kilimanjaro to the Earth’s horizon.2 Make a PlanLet C be the center of the Earth, E be the summit of Mt. Kilimanjaro, and H be a point on the horizon. Find the length of

−− EH , which is tangent to circle C at H. By Thm. 11-1-1,

−− EH ⊥ −− CH .

So �CHE is a right �.3 SolveED = 19,340 ft

= 19,340

______ 5280

≈ 3.66 mi

EC = CD + ED ≈ 4000 + 3.66 = 4003.66 mi E C 2 ≈ E H 2 + C H 2 4003.6 6 2 ≈ E H 2 + 400 0 2 29,293.40 ≈ E H 2 171 mi ≈ EH4 Look BackThe problem asks for the distance to the nearest mile. Check that the answer is reasonable by using the Pythagorean Thm. Is 17 1 2 + 400 0 2 ≈ 400 4 2 ? Yes, 16,029,241 ≈ 16,032,016.

4a. By Thm. 11-1-3, RS = RT x __

4 = x - 6.3

x = 4x - 25.2-3x = -25.2 x = 8.4

RS = (8.4)

____ 2 = 2.1

b. By Thm.11-1-3, RS = RTn + 3 = 2n - 1 3 = n - 1 4 = n RS = (4) + 3 = 7

THINK AND DISCUSS, PAGE 750

1. 4 lines

2. No; if line is tangent to the circle with the larger radius, it will not intersect the circle with the smaller radius. If the line is tangent tothe circle with the smaller radius, it will intersect the circle with the larger radius at 2 points.

3. No; a circle consists only of those points which are a given distance from the center.

4. By Thm. 11-1-1, m∠PQR = 90°. So by Triangle Sum Theorem m∠PRQ = 180 - (90 + 59) = 31°.

Copyright © by Holt, Rinehart and Winston. 273 Holt GeometryAll rights reserved.

5.

EXERCISES, PAGES 751–754

GUIDED PRACTICE, PAGE 751

1. secant 2. concentric

3. congruent

4. chord: −− EF ; tangent: m; radii:

−− DE , −− DF ; secant: �;

diameter: −− EF

5. chord: −− QS ; tangent: � �� ST ; radii:

−− PQ , −− PR ,

−− PS ;

secant: � �� QS ; diameter: −− QS

6. radius of circle A: 4 - 1 = 3radius of circle B: 4 - 2 = 2point of tangency: (-1, 4)equation of tangent line: y = 4

7. radius of circle R: 4 - 2 = 2radius of circle S: 4 - 2 = 2point of tangency: (1, 2)equation of tangent line: x = 1

8. 1 Understand the ProblemThe answer will be the length of an imaginary segment from the ISS to the Earth’s horizon.2 Make a PlanLet C be the center of the Earth, let E be ISS, and let H be the point on the horizon. Find the length of

−− EH , which is tangent to circle C at H. By Thm. 11-1-1, −− EH ⊥

−− CH . So CHE is a right �.3 SolveEC = CD + ED ≈ 4000 + 240 = 4240 mi E C 2 ≈ E H 2 + C H 2 424 0 2 ≈ E H 2 + 400 0 2 1,977,60 ≈ E H 2 1406 mi ≈ EH4 Look BackThe problem asks for the distance to the nearest mile. Check that answer is reasonable by using the Pythagorean Thm. Is 1406 2 + 400 0 2 ≈ 4240 2 ? Yes, 17,976,836 ≈ 17,977,600.

9. By Thm. 11-1-3, JK = JL4x - 1 = 2x + 92x - 1 = 9 2x = 10 x = 5JK = 4(5) - 1 = 19

10. By Thm. 11-1-3, ST = SU y - 4 = 3 _

4 y

4y - 16 = 3y y - 16 = 0 y = 16ST = (16) - 4 = 12

PRACTICE AND PROBLEM SOLVING, PAGES 752–754

11. chords: −− RS ,

−−− VW ; tangent: �; radii: −− PV ,

−−− PW ;

secant: � �� VW ; diameter: −−− VW

12. chords: −− AC ,

−− DE ; tangent: � �� CF ; radii: −− BA ,

−− BC ;

secant: � �� DE ; diameter: −− AC

13. radius of circle C: 2 - 0 = 2radius of circle D: 4 - 0 = 4point of tangency: (-4, 0)equation of tangent line: x = -4

14. radius of circle M: 3 - 2 = 1radius of circle N: 5 - 2 = 3point of tangency: (2, 1)equation of tangent line: y = 1

15. 1 Understand the ProblemThe answer will be the length of an imaginary segment from the summit of Olympus Mons to Mars’ horizon.2 Make a PlanLet C be the center of Mars, let E be summit of Olympus Mons, and let H be a point on the horizon. Find the length of

−− EH , which is tangent to circle C at H. By Thm. 11-1-1,

−− EH ⊥ −− CH .

So triangle CHE is a right triangle.3 SolveEC = CD + ED

≈ 3397 + 25 = 3422 km E C 2 ≈ E H 2 + C H 2 3422 2 ≈ E H 2 + 3397 2 170,475 ≈ E H 2 413 km ≈ EH4 Look BackThe problem asks for the distance to the nearest km. Check that the answer is reasonable by using the Pythagorean Thm. Is 413 2 + 3397 2 ≈ 3422 2 ? Yes, 11,710,178 ≈ 11,710,084.

16. By Thm. 11-1-3, AB = AC2 x 2 = 8xSince x ≠ 0, 2x = 8 x = 4AB = 2(4 ) 2 = 32

17. By Thm. 11-1-3,RS = RT

y = y 2

___ 7

7y = y 2 Since y ≠ 0, 7 = y

RT = (7 ) 2

____ 7 = 7

18. S (true if circles are identical)


19. N 20. N

21. A

22. S (if chord passes through center)

23. −− AC 24.

−− PA , −− PB ,

−− PC , −− PD

25. −− AC

26. By Thm. 11-1-1, Thm 11-1-3, the definition of a circle, and SAS, �PQR � �PQS; so

�� PQ bisects

∠RPS. Therefore, m∠QPR = 1 _ 2 (42) = 21°.

By Thm 11-1-1, �PQR is a right �. By the Sum, m∠PQR + m∠PRQ + m∠QPR = 180 m∠PQR + 90 + 21 = 180 m∠PQR = 180 - (90 + 21) = 69°m∠PQS = 2m∠PQR = 2(69) = 138°

27. By Thm 11-1-1, m∠R = m∠S = 90°.By Quad. Sum Thm.,m∠P + m∠Q + m∠R + m∠S = 360 x + 3x + 90 + 90 = 360 4x + 180 = 360 4x = 180 x = 45m∠P = 45°

28a. The perpendicular segment from a point to a line is the shortest segment from the point to the line.

b. B c. radius

d. line � ⊥ −− AB

29. Let E be any point on the line m other than D. It is given that line m ⊥

−− CD . So �CDE is a right � with hypotenuse

−− CE . Therefore, CE > CD. Since −− CD is

a radius, E must lie in exterior of circle C. Thus D is only a point on the line m that is also on circle C. So the line m is tangent to circle C.

30. Since 2 points determine a line, draw auxiliary segments

−− PA , −− PB , and

−− PC . Since −− AB and

−− AC are tangents to circle P,

−− AB ⊥ −− PB and

−− AC ⊥ −− PC . So

�ABP and �ACP are right �. −− PB �

−− PC since they are both radii of circle P, and

−− PA � −− PA by Reflex.

Prop. of �. Therefore, �ABP � � ACP by HL � and

−− AB � −− AC by CPCTC.

31. QR = QS = 5 QT 2 = QR 2 + RT 2 (ST + 5 ) 2 = 5 2 + 12 2 ST + 5 = 13 ST = 8

32. AB = AD 23 = x AC = AE 23 + x - 5 = x + DE23 + 23 - 5 = 23 + DE 41 = 23 + DE DE = 18

33. JK = JL and JL = JM, so, JK = JM JK = JM6y - 2 = 30 - 2y 8y = 32 y = 4JL = JM = 30 - 2(4) = 22

34. Point of tangency must be (x, 2), where x - 2 = ±3x = 5 or -1.Possible points of tangency are (5, 2) and (-1, 2).

35a. BCDE is a rectangle; by Thm. 11-1-1, ∠BCD and ∠EDC are right . It is given that ∠DEB is a right ∠. ∠CBE must also be a right by Quad. Sum Thm. Thus, BCDE has 4 right and is a rectangle.

b. BE = CE = 17 in.AE = AD - DE = AD - BC = 5 - 3 = 2 in.

c. AB 2 = AE 2 + BE 2 = 2 2 + 17 2 AB 2 = 293 AB = √ �� 293 ≈ 17.1 in.

36. Not possible; if it were possible, �XBC would contain 2 right . which contradicts � Sum Thm.

37. By Thm 11-1-1, ∠R and ∠S are right . By Quad. Sum Thm.,∠P + ∠Q + ∠R + ∠S = 360 ∠P + ∠Q + 90 + 90 = 360 ∠P + ∠Q = 180By definition, ∠P and ∠Q are supplementary angles.

TEST PREP, PAGE 754

38. CA D 2 = A B 2 + B D 2 = 10 2 + 3 2 = 109 AD = √ �� 109 ≈ 10.4 cm

39. G-2 - (-4) = 2. So, (3, -4) lies on circle P; y = -4 meets circle P only at (3, -4). So it is tangent to circle P.

40. B

π(5 ) 2

_____ π(6 ) 2

= 25π

____ 36π

= 25 ___ 36

CHALLENGE AND EXTEND, PAGE 754

41. Since 2 points determine a line, draw auxiliary segments

−− GJ and −− GK . It is given that

−− GH ⊥ −− JK , so,

∠GHJ and ∠GHK are right . Therefore, �GHJ and �GHK are right .

−− GH � −− GH by Reflex. Prop.

of �, and −− GJ �

−− GK because they are radii of circle G. Thus �GHJ � �GHK by HL, and

−− JH � −− KH by

CPCTC.

42. By Thm. 11-1-1, ∠C and ∠D are right . So BCDE is a rectangle, CE = DB = 2, and BE = DC = 12. Therefore, �ABE is a right with leg lengths 5 - 2 = 3 and 12. So

AB = √ �� A E 2 + B E 2 =

√ �� 3 2 + 1 2 2 = √ �� 153 = 3 √ �� 17

43. Draw a segment from X to the center C of the wheel. ∠XYC is a right angle and m∠YXC = 1 _

2 (70) = 35°. So

tan 35° = 13 ___ XY

XY = 13 ______ tan 35° ≈ 18.6 in.


SPIRAL REVIEW, PAGE 754

44. 14 + 6.25h > 12.5 + 6.5h 1.5 > 0.25h 6 > hSince h is positive, 0 < h < 6.

45. P = LM + PR ________ LR

= 10 + (16 + 4)

_________________ 10 + 6 + 4 + 16 + 4

= 30 ___ 40

= 3 __ 4

46. P = LP ___ LR

= 10 + 6 + 4 __________ 40

= 20 ___ 40

= 1 __ 2

47. P = MN + PR ________ LR

= 6 + (16 + 4)

___________ 40

= 26 ___ 40

= 13 ___ 20

48. P = QR ___ LR

= 4 ___ 40

= 1 ___ 10

CONNECTING GEOMETRY TO DATA ANALYSIS: CIRCLE GRAPHS, PAGE 755

TRY THIS, PAGE 755

1. Step 1 Add all the amounts.18 + 10 + 8 = 36Step 2 Write each part as a fraction of the whole.novels: 18 __

36 reference: 10 __

36 textbooks: 8 __

36

Step 3 Multiply each fraction by 360° to calculate central ∠ measure.novels: 18 __

36 (360) = 180° reference: 10 __

36 (360) = 100°

textbooks: 8 __ 36

(360) = 80°Step 4 Match a circle graph to the data.The data match graph D.

2. Step 1 Add all the amounts.450 + 120 + 900 + 330 = 1800Step 2 Write each part as a fraction of the whole.travel: 450 ____

1800 meals: 120 ____

1800 lodging: 900 ____

1800

other: 330 ____ 1800

Step 3 Multiply each fraction by 360° to calculate central ∠ measure.travel: 450 ____

1800 (360) = 90° meals: 120 ____

1800 (360) = 24°

lodging: 900 ____ 1800

(360) = 180° other: 330 ____ 1800

(360) = 66°Step 4 Match a circle graph to the data.The data match graph C.

3. Step 1 Add all the amounts.190 + 375 + 120 + 50 = 735Step 2 Write each part as a fraction of the whole.food: 190 ___

735 health: 375 ___

735 training: 120 ___

735 other: 50 ___

735

Step 3 Multiply each fraction by 360° to calculate central ∠ measure.food: 190 ___

735 (360) ≈ 93° health: 375 ___

735 (360) ≈ 184°

training: 120 ___ 735

(360) ≈ 59° other: 50 ___ 735

(360) ≈ 24°Step 4 Match a circle graph to the data.The data match graph B.

11-2 ARCS AND CHORDS, PAGES 756–763


1a. m∠FMC = (0.03 + 0.09 + 0.10 + 0.11)360° = 108°

b. m∠ � AHB = (1 - 0.25)360° = 270° c. m∠EMD = (0.10)360° = 36° 2a. m∠JPK = 25° (Vert. � Thm.)

m � JK = 25°m∠KPL + m∠LPM + m∠MPN = 180° m∠KPL + 40° + 25° = 180° m∠KPL = 115° m � KL = 115°m � JKL = m � JK + m � KL

= 25° + 115° = 140°

b. m � LK = m � KL = 115°m∠KPN = 180°m � KJN = 180°m � LJN = m � LK + m � KJN

= 180° + 115° 3a. m∠RPT = m∠SPT

RT = ST 6x = 20 - 4x 10x = 20 x = 2RT = 6(2) = 12

b. m∠CAD = m∠EBF (11-2-2(3)) mCD = mEF 25y = 30y - 20 20 = 5y y = 4mCD = m∠CAD = 25(4) = 100°

4. Step 1 Draw radius −− PQ .

PQ = 10 + 10 = 20Step 2 Use Pythagorean and 11-2-3.P T 2 + Q T 2 = P Q 2 1 0 2 + Q T 2 = 2 0 2 Q T 2 = 300 QT = √ �� 300 = 10 √ � 3 Step 3 Find QR.QR = 2 (10 √ � 3 ) = 20 √ � 3 ≈ 34.6


ge07_SOLKEY_C11_273-300.indd 276ge07_SOLKEY_C11_273-300.indd 276 10/2/06 4:29:41 PM10/2/06 4:29:41 PM


1. The arc measures between 90° and 180°. 2. if arcs are on 2 different circles with different radii

3.



1. semicircle 2. Vertex is the center of the circle.

3. major arc 4. minor arc

5. m∠PAQ = 0.45(360) = 162° 6. m∠VAU = 0.07(360) = 25.2° 7. m∠SAQ = (0.06 + 0.11)360 = 61.2°

8. m � UT = m∠UAT = 0.1(360) = 36°

9. m � RQ = m∠RAQ = 0.11(360) = 39.6°

10. m � UPT = (1 - 0.1)360 = 324°

11. m � DE = m∠DAE = 90°m � EF = m∠EAF = m∠BAC = 90 - 51 = 39°m � DF = m � DE + m � EF = 90 + 39 = 129°

12. m � DEB = m∠DAE + m∠EAB = 90 + 180 = 270°

13. m∠HGJ + m∠JGL = m∠HGL 72 + m∠JGL = 180 m∠JGL = 108° m � JL = 108°

14. m � HLK = m∠HGL + m∠LGK = 180 + 30 = 210° 15. QR = RS (Thm. 11-2-2(1))

8y - 8 = 6y 2y = 8 y = 4QR = 8(4) - 8 = 24

16. m∠CAD = m∠EBF (Thm. 11-2-2(3)) 45 - 6x = -9x 3x = -45 x = -15m∠EBF = -9(-15) = 135°

17. Step 1 Draw radius −− PR .

PR = 5 + 8 = 13Step 2 Use the Pythagorean Thm. and Thm. 11-2-3.Let the intersection of

−− PQ and −− RS be T.

P T 2 + R T 2 = P R 2 5 2 + R T 2 = 13 2 RT = 12Step 3 Find RS.RS = 2(12) = 24

18. Step 1 Draw radius −− CE .

CE = 50 + 20 = 70Step 2 Use the Pythagorean Thm. and Thm. 11-2-3.Let the intersection of

−− CD and −− EF be G.

C G 2 + E G 2 = C E 2 5 0 2 + E G 2 = 7 0 2 RG = √ �� 2400 = 20 √ � 6 Step 3 Find EF.EF = 2 (20 √ � 6 ) = 40 √ � 6 ≈ 98.0


19. m∠ADB = 35 ____________ 35 + 39 + 29

(360) = 35 ____ 103

(360) ≈ 122.3°

20. m∠ADC = 29 ____ 103

(360) = 101.4°

21. m � AB = m∠ADB ≈ 122.3°

22. m � BC = m∠BDC = 39 ____ 103

(360) ≈ 136.3°

23. m � ACB = 360 - m∠ADB ≈ 360 - 122.3 = 237.7°

24. m � CAB = 360 - m∠BDC ≈ 360 - 136.3 = 223.7°

25. m � MP = m∠MJP = m∠MJQ - m∠PJQ = 180 - 28 = 152°

26. mQNL = mQNM + mML= m∠QJM + m∠MJL= 180 + 28 = 208°

27. m � WT = m � WS + m � ST = m∠WXS + m∠SXT= 55 + 100 = 155°

28. m � WTV = m � WS + m � STV = m∠WXS + m∠SXV= 55 + 180 = 235°

29. m∠CAD = m∠EBF (Thm. 11-2-2(3))10x - 63 = 7x 3x = 63 x = 21m∠CAD = 10(21) - 63 = 147°

30. m � JK = m � LM (Thm. 11-2-2(2))4y + y = y + 68 y = 17m � JK = 4(17) + 17 = 85°


31. AC = AB = 2.4 + 1.7 = 4.1Let

−− AB and −− CD meet at E.

A E 2 + CE 2 = AC 2 2. 4 2 + C E 2 = 4. 1 2 CE = √ �� 11.05 CD = 2 √ �� 11.05 ≈ 6.6

32. PR = PQ = 2(3) = 6Let

−− PQ and −− RS meet at T.

P T 2 + R T 2 = P R 2 3 2 + R T 2 = 6 2 RT = √ �� 27 = 3 √ � 3 RS = 2 (3 √ � 3 ) = 6 √ � 3 ≈ 10.4

33. F; the ∠ measures between 0° and 180°. So it could be right or obtuse.

34. F; Endpts. of a diameter determine 2 � arcs measuring exactly 180°.

35. T (Thm. 11-2-4)

36. Check students’ graphs.

37. Let m∠AEB = 3x, m∠BEC = 4x, and m∠CED = 5x.m∠AEB + m∠BEC + m∠CED = 180 3x + 4x + 5x = 180 12x = 128 x = 15m∠AEB = 3(15) = 45° m∠BEC = 4(15) = 60°m∠CED = 5(15) = 75°

38. m � JL + m � JK + m � KL = 3607x - 18 + 4x - 2 + 6x + 6 = 360 17x = 374 x = 22m � JL = 7(22) - 18 = 136°

39. m∠QPR = 180 10x = 180 x = 18m∠SPT = 6(18) = 108°

40. Statements Reasons

1. −− BC �

−− DE 1. Given.2.

−− AB � −− AD and

−− AC � −− AE

2. All radii of a circle are �.

3. BAC � DAE 3. SSS4. ∠BAC � ∠DAE 4. CPCTC5. m∠BAC = m∠DAE 5.

6. m � BC = m � DE 6. Definition of arc measures

7. � BC � � DE 7. Definition of arcs


1. � BC � � DE 1. Given

2. m � BC = m � DE 2. Definition of � arcs3. m∠BAC = m∠DAE 3. Definition of arc

measures4. ∠BAC � ∠DAE 4.


1. −− CD ⊥

−− EF 1. Given 2. Draw radii

−− CE and

−− CF . 2. 2 points determine

a line. 3.

−− CE � −− CF 3. All radii of a circle

are congruent. 4.

−−− CM � −−− CM 4. Reflex. Prop. of �

5. ∠CMF and ∠CME are rt. �.

5. Def. of ⊥

6. CMF and CME are rt. �.

6. Def. of a rt.

7. CMF � CME 7. HL Steps 3, 4 8.

−− FM � −− EM 8. CPCTC

9. −− CD bisects

−− EF 9. Def. of a bisector10. ∠FCD � ∠ECD 10. CPCTC11. m∠FCD = m∠ECD 11. Def. of � �12. m � FD = m � ED 12. Def. of arc measures

13. � FD � � ED 13. Def. of arcs14.

−− CD bisects � EF . 14. Def. of a bisector


1. −− JK is the ⊥ bis. of

−− GH 1. Given

2. A is equidistant from G and H.

2. Def. of the center of circle

3. A lies on the ⊥ bis. of

−− GH .3. Perpendicular

Bisector Theorem4.

−− JK is a diameter of circle A.

4. Def. of diam.

44. The circle is divided into eight � sectors, each with central ∠ measure 45°. So possible measures of the central congruent are multiples of 45° between 0(45) = 0° and 8(45) = 360°. So there are three different sizes of angles: 135°, 90°, and 45°.

45. Solution A is incorrect because it assumes that ∠BGC is a right ∠.

46. To make a circle graph, draw a circle and then draw central � that measure 0.4(360) = 144°, 0.35(360) = 126°, 0.15(360) = 54°, and 0.1(360) = 36°.

47a. AC = 1 _ 2 (27) = 13.5 in.

AD = AB - DB = 13.5 - 7 = 6.5 in.

b. C D 2 + A D 2 = A C 2 C D 2 + 6. 5 2 = 13. 5 2 CD = √ �� 140 = 2 √ �� 35 ≈ 11.8 in.

c. By Theorem 11-2-3, −− AB bisects

−− CE . So CE = 2CD = 4

√ �� 35 ≈ 23.7 in.

TEST PREP, PAGE 763

48. Dm � WT = 90 + 18 = 108°m � UW = 90°

m � VR = 180 - 41 = 139°m � TV = 180 - 18 = 162°

49. FCE = 1 _

2 (10) = 5

A E 2 + 5 2 = 6 2 AE = √ � 11 ≈ 3.3



50. 90 −− AP is a horizontal radius, and

−− BP is a vertical radius. So m � AB = m∠APB = 90°.


51. AD = AB = 4 + 2 = 6cos BAD = 4 _

6 = 2 _

3

m � BD = m∠BAD = co s -1 ( 2 _ 3 ) ≈ 48.2°

52. 2 points determine 2 distinct arcs.3 points determine 6 arcs.4 points determine 12 arcs.5 points determine 20 arcs.�n points determine n(n - 1) arcs.

53a. π → 180°. So π __ 2 → 90°, π

__ 3 → 60°, and π

__ 4 → 45°

b. 135° → 135 ____ 180

(π) = 3π

___ 4

270° → 270 ____ 180

(π) = 3π

___ 2


54. (3x ) 3 (2 y 2 ) ( 3 -2 y 2 )

(27 x 3 ) (2 y 2 ) ( 1 _ 9 y 2 )

6 x 3 y 4

55. a 4 b 3 (-2a) -4

a 4 b 3 ( 1 __ 16

a -4 )

1 __ 16

b 3

56. (-2 r 3 s 2 ) (3t s 2 ) 2

-2 t 3 s 2 (9 t 2 s 4 ) -18 t 5 s 6

57. 3 = 1 + 2 7 = 3 + 413 = 7 + 621 = 13 + 821 + 10 = 31

58. C, E, G, I, K, M 59. 6 = 1 + 515 = 6 + 915 + 13 = 28

60. ∠NPQ and ∠NMQ are right � (Thm. 11-1-2).So ∠NMQ = 90°.

61. PQ = MQ (Thm. 11-1-3) 2x = 4x - 9 9 = 2x x = 4.5MQ = 4(4.5) - 9 = 9

CONSTRUCTION, PAGE 763

1. O is on the ⊥ bisector of −− PQ . So by Conv. of ⊥

Bisector, OP = OQ. Similarly, O is on ⊥ bisector of

−− QR . So OQ = OR. Thus, −− OP ,

−−− OQ , and −− OR are

radii of circle O, and circle O contains Q and R.

11-3 SECTOR AREA AND ARC LENGTH, PAGES 764–769


1a. A = π r 2 ( m ____ 360

) = π(1 ) 2 ( 90 ____ 360

) = 1 __ 4

π m 2 ≈ 0.79 m 2

b. A = π r 2 ( m ____ 360

) = π(16 ) 2 ( 36 ____ 360

) = 25.6π in. 2 ≈ 80.42 in. 2

2. A = π r 2 ( m ____ 360

)

= π(360 ) 2 ( 180 ____ 360

)

≈ 203,575 ft 2

3. Step 1 Find the area of sector RST.

A = π r 2 ( m ____ 360

) = π(4 ) 2 ( 90 ____ 360

) = 4π m 2

Step 2 Find the area of RST.A = 1 _

2 bh = 1 _

2 (4)(4) = 8 m 2

Step 3 area of segment = area of sector RST - area of

RST= 4π - 8 ≈ 4.57 m 2

4a. L = 2πr ( m ____ 360

) = 2π(6) ( 40 ____ 360

) = 4 __ 3

π m ≈ 4.19 m

b. L = 2πr ( m ____ 360

) = 2π(4) ( 135 ____ 360

) = 3π cm ≈ 9.42 cm


1. An arc measure is measured in degrees. An arc length is measured in linear units.

2. the radius and central ∠ of the sector

3.



1. segment

2. A = π r 2 ( m ____ 360

) = π(6 ) 2 ( 90 ____ 360

) = 9π m 2 ≈ 28.27 m 2

3. A = π r 2 ( m ____ 360

) = π(8 ) 2 ( 135 ____ 360

) = 24π cm 2 ≈ 75.40 cm 2

4. A = π r 2 ( m ____ 360

) = π(2 ) 2 ( 20 ____ 360

) = 2 __ 9

π ft 2 ≈ 0.70 ft 2

5. A = π r 2 ( m ____ 360

) = π(3 ) 2 ( 150 ____ 360

) = 15 ___ 4 π mi 2 ≈ 12 mi 2



6. Step 1 Find area of sector ABC.

A = π r 2 ( m ____ 360

) = π(3 ) 2 ( 90 ____ 360

) = 9 __ 4 π in. 2

Step 2 Find area of �ABC.A = 1 _

2 bh = 1 _

2 (3)(3) = 9 _

2 in. 2

Step 3 area of segment = area of sector ABC - area of

�ABC= 9 _

4 π - 9 _

2 ≈ 2.57 in. 2

7. Step 1 Find the area of sector DEF.

A = π r 2 ( m ____ 360

) = π(20 ) 2 ( 60 ____ 360

) = 200 ____ 3 π m 2

Step 2 Find the area of �DEF.A = 1 _

2 bh = 1 _

2 (20) (10 √ � 3 ) = 100 √ � 3 m 2

Step 3 area of segment = area of sector DEF - area of

�DEF= 200 ___

3 π - 100 √ � 3 ≈ 36.23 m 2

8. Step 1 Find the area of sector ABC.

A = π r 2 ( m ____ 360

) = π(6 ) 2 ( 45 ____ 360

) = 9 __ 2 π cm 2

Step 2 Find the area of �ABC.A = 1 _

2 bh = 1 _

2 (6 ) (3 √ � 2 ) = 9 √ � 2 cm 2

Step 3 area of segment = area of sector ABC - area of

�ABC = 9 __

2 π - 9 √ � 2 ≈ 1.41 cm 2

9. L = 2πr ( m ____ 360

) = 2π(16) ( 45 ____ 360

) = 4π ft ≈ 12.57 ft

10. L = 2πr ( m ____ 360

) = 2π(9) ( 120 ____ 360

) = 6π m ≈ 18.85 m

11. L = 2πr ( m ____ 360

) = 2π(6) ( 20 ____ 360

) = 2 __ 3 π in. ≈ 2.09 in.


12. A = π(20 ) 2 ( 150 ____ 360

) = 500 ____ 3 π m 2 ≈ 523.60 m 2

13. A = π(9 ) 2 ( 100 ____ 360

) = 45 ___ 2 π in 2 ≈ 70.69 in. 2

14. A = π(2 ) 2 ( 47 ____ 360

) = 47 ___ 90

π ft 2 ≈ 1.64 ft 2

15. A = π(20 ) 2 ( 180 ____ 360

) = 200π ≈ 628 in. 2

16. �ABC is a 45°-45°-90° triangle. So central angle measures 90°.area of segment = area of sector ABC - area of

�ABC

= π(10 ) 2 ( 90 ____ 360

) - 1 __ 2 (10)(10)

= 25π - 50 ≈ 28.54 m 2

17. m∠KLM = m � KM = 120°area of segment = area of sector KLM - area of

�KLM

= π(5 ) 2 ( 120 ____ 360

) - 1 __ 2 ( 5 __

2 ) (5 √ � 3 )

= 25 ___ 3 π - 25 ___

4 √ � 3 ≈ 15.35 in. 2

18. area of segment = area of sector RST - area of �RST

= π(1 ) 2 ( 60 ____ 360

) - 1 __ 2 (1) (

√ � 3 ___

2 )

= 1 __ 6 π - 1 __

4 √ � 3 ≈ 0.09 in. 2

19. L = 2π(5) ( 50 ____ 360

) = 25 ___ 18

π mm ≈ 4.36 mm

20. L = 2π(1.5) ( 160 ____ 360

) = 4 __ 3 π m ≈ 4.19 m

21. L = 2π(2) ( 9 ____ 360

) = 1 ___ 10

π ft ≈ 0.31 ft

22. P = 2π(3) ( 180 ____ 360

) + 3 (2π(1) ( 180 ____ 360

) ) = 6π ≈ 18.8 in.

23. never

24. sometimes (if radii of arcs are equal)

25. always

26. A = π r 2 ( m ____ 360

)

9π = π r 2 ( 90 ____ 360

)

36 = r 2 r = 6

27. L = 2πr ( m ____ 360

)

8π = 2πr ( 120 ____ 360

)

24π = 2πr r = 12

28a. L ≈ 2 ( 22 ___ 7 ) (7) ( 90 ____

360 ) = 11 in.

b. L = 2π(7) ( 90 ____ 360

) = 7 __ 2 π ≈ 10.99557429 in.

c. overestimate, since L < 10.996 < 11

29a. L = 2π(2.5) ( 90 ____ 360

) = 5 __ 4 π ≈ 3.9 ft

b. 4.5 = 2π(2.5) ( m ____ 360

)

9 ____ 10π

= m ____ 360

m = 324 ____ π

≈ 103°

30. The area of sector BAC is 45 ____ 360

= 1 __ 8 area of circle A.

So if area of circle A is 24 in 2 , the area of the

sector will automatically be 1 __ 8 (24) = 3 in. 2

So we need only to solve π r 2 = 24 for r.π r 2 = 24

r 2 = 24 ___ π

r = √ �� 24 ___ π

≈ 2.76 in.

31. If the length of the arc is L and its degree measure is m, then L = 2πr ( m ____

360 )

360L = 2πrm

r = 360L _____ 2πm

= 180L _____ πm .


TEST PREP, PAGE 769

32. BA = π(8 ) 2 ( 90 ____

360 ) = 16π

33. GA = 2π(8) ( 90 ____

360 ) = 4π

34. 43.98A = π(12 ) 2 ( 35 ____

360 ) ≈ 43.98


35. A = π(5 ) 2 ( 40 ____ 360

) - π(2 ) 2 ( 40 ____ 360

) = 25 ___ 9 π - 4 __

9 π = 7 __

3 π

36a. V = Bh = (π(4 ) 2 ( 30 ____ 360

) ) (3) = 4π ≈ 12.6 in. 3

b. B = 2 (π(4 ) 2 ( 30 ____ 360

) ) = 8 __ 3 π

L = 2(3)(4) + 2 (π(4) ( 30 ____ 360

) ) (3) = 24 + π

A = 8 __ 3 π + 24 + 2π ≈ 38.7 in. 2

37a. A(�) = π(2 ) 2 = 4π

A(red) = 4π(1 ) 2 ( 45 ____ 360

) = 1 __ 2 π

P(red) = 1 _ 2 π

___ 4π

= 1 __ 8

b. A(blue) = 4 (π(2 ) 2 ( 45 ____ 360

) - π(1 ) 2 ( 45 ____ 360

) ) = 3 __ 2 π

P(blue) = 3 _ 2 π

___ 4π

= 3 __ 8

c. P(red or blue) = 1 _ 2 π + 3 _

2 π

________ 4π

= 1 __ 2


38. 8x - 2y = 6 2y = 8x - 6 y = 4x - 3The line is ‖.

39. slope = 2 - 0 ______ 1 1 _

2 - 1 _

2 = 2

The line is neither ‖ nor ⊥.

40. y = mx + 1 0 = m(4) + 1

m = - 1 __ 4

The line is ⊥.

41. V = 4 _ 3 π(3 ) 3 = 36π cm 3

42. S = 4π = 4π r 2 r 2 = 1 r = 1 cmC = 2π(1) = 2π cm

43. m∠KLJ = m∠KLH - (m∠GLJ - m∠GLH)10x - 28 = 180 - (90 - (2x + 2))10x - 28 = 90 + 2x + 2 8x = 120 x = 15m∠KLJ = 10(15) - 28 = 122°

44. m � KJ = m∠KLJ = 122°

45. m � JFH = m � JF + m � FG + m � GH = m∠JLF + m∠FLG + m∠GLH= 180 + 90 + 2(15) + 2 = 302°

11A MULTI-STEP TEST PREP, PAGE 770

1. Let � represent the length of the stand. Radius r = 14 in. Form a right � with leg lengths 14 - 10 = 4 in. and 1 _

2 � in., and

hypotenuse length 14 in.

4 2 + ( 1 _ 2 �)

2 = 14 2

16 + 1 _ 4 � 2 = 196

1 _ 4 � 2 = 180

� 2 = 720 � = √ 720 = 12 √ 5 ≈ 26.8 in.

2. Let E be on −− AD with

−− BE −− CD and BE = CD.

�ABE is a right �. So A E 2 + BE 2 = AB 2 (4 - 2 ) 2 + CD 2 = 15 2 C D 2 = 221 CD = √ 221 ≈ 14.9 in.

3. L = 2πr ( m ____ 360

) = 2π(4) ( 60 ____ 360

) = 4 __ 3

π ≈ 4.2 in.

4. L = 2πr ( m ____ 360

)

4 __ 3 π = 2π(2) ( m ____

360 )

1 __ 3 = m ____

360

m = 360 ____ 3 = 120°

11A READY TO GO ON? PAGE 771

1. chord: −− PR ; tangent: m; radii:

−− QP , −− QR ,

−− QS ; secant: � � PR ; diameter:

−− PR

2. chords: −− BD ,

−− BE ; tangent: � � BC ; radii: −− AB ,

−− AE ; secant: � � BD ; diameter:

−− BE

3. Let x be the distance to the horizon. The building forms a right � with leg lengths x mi and 4000 mi, and hypotenuse length (4000 + 732 ____

5280 ) mi.

x 2 + (4000 ) 2 = (4000 + 732 ____ 5280

) 2

x 2 ≈ 1109.11 x ≈ 33 mi

4. m � BC + m � CD = m � BD m � BD + m∠CAD = m∠BAD m � BD + 49 = 90 m � BD = 41°

5. m � BED = m � BFE + m � ED = m∠BAE + m∠EAD= 180 + 90 = 270°

6. m � SR + m � RQ = m � SQ m � SR + m∠RPQ = m∠SPQ m � SR + 71 = 180 m � SR = 109°



7. m � SQU + m � UT + m � TS = 360m � SQU + m∠UPT + m∠TPS = 360 m � SQU + 40 + 71 = 360 m � SQU = 249°

8. Let JK = 2x; then x 2 + 4 2 = (4 + 3 ) 2 x 2 = 33 x = √ �� 33 JK = 2 √ �� 33 ≈ 11.5

9. Let XY = 2x; then x 2 + 4 2 = 8 2

x 2 = 48 x = 4 √ � 3 XY = 8 √ � 3 ≈ 13.9

10. A = π(22 ) 2 ( 80 ____ 360

) = 968 ____ 9 π ≈ 338 cm 2

11. � AB = 2π(4) ( 150 ____ 360

) = 10 ___ 3 π ft ≈ 10.47 ft

12. � EF = 2π(2.4) ( 75 ____ 360

) = π cm ≈ 3.14 cm

13. arc length = 2π(5) ( 44 ____ 360

) = 11 ___ 9 π in. ≈ 3.84 in.

11. arc length = 2π(46) ( 180 ____ 360

) = 46π m ≈ 144.51 m

11-4 INSCRIBED ANGLES, PAGES 772–779


1a. m∠ABC = 1 _ 2 m � ADC

135 = 1 _ 2 m � ADC

m � ADC = 270°

b. m∠DAE = 1 _ 2 mDE

= 1 _ 2 (72) = 36°

2. m∠ABD = 1 _ 2 m � AD = 1 _

2 (86) = 43°

m∠BAC = 1 _ 2 m � BC

60 = 1 _ 2 m � BC

m � BC = 120° 3a. ∠ABC is a right angle

m∠ABC = 90 8z - 6 = 90 8z = 96 z = 12

b. m∠EDF = m∠EGF 2x + 3 = 75 - 2x 4x = 72 x = 18m∠EDF = 2(18) + 3

= 39° 4. Step 1 Find the value of x.

m∠K + m∠M = 18033 + 6x + 4x - 13 = 180 10x = 160 x = 16Step 2 Find the measure of each ∠.m∠K = 33 + 6(16) = 129°m∠L =

9(16) ____

2 = 72°

m∠M = 4(16) - 13 = 51°m∠J + m∠L = 180 m∠J + 72 = 180 m∠J = 108°


1. No; a quadrilateral can be inscribed in a circle if and only if its opposite are supplementary.

2. An arc that is 1 _ 4 of a circle measures 90°. If the arc

measures 90°, then the measure of the inscribed ∠ is 1 _ 2 (90) = 45°.

3.



1. inscribed

2. m∠DEF = 1 _ 2 m � DF

= 1 _ 2 (78) = 39°

3. m∠EFG = 1 _ 2 m � EG

29 = 1 _ 2 m � EG

m � EG = 58°

4. m∠JNL = 1 _ 2 m � JKL

102 = 1 _ 2 m � JKL

m � JKL = 204°

5. m∠LKM = 1 _ 2 m � LM

= 1 _ 2 (52) = 26°

6. m∠QTR + m∠Q + m∠R = 180m∠QTR + 1 _

2 m � RS + m∠S = 180

m∠QTR + 1 _ 2 (90) + 25 = 180

m∠QTR + 70 = 180 m∠QTR = 110°

7. ∠DEF is a right ∠ 4x ___ 5 = 90

4x = 450 x = 112.5

8. ∠FHG is a right ∠. So �FGH is a 45°-45°-90° triangle.m∠GFH = 45 3y + 6 = 45 3y = 39 y = 13

9. m∠XYZ = m∠XWZ 7y - 3 = 4 + 6y y = 7m∠XYZ = 7(7) - 3 = 46°


10. Step 1 Find the value of x. m∠P + m∠R = 1805x + 20 + 7x - 8 = 180 12x = 168 x = 14Step 2 Find the measures.m∠P = 5(14) + 20 = 90°m∠Q = 10(14) = 140°m∠R = 7(14) - 8 = 90°m∠S + m∠Q = 180 m∠S + 140 = 180 m∠S = 40°

11. Step 1 Find the value of z. m∠A + m∠C = 1804z - 10 + 10 + 5z = 180 9z = 180 z = 20Step 2 Find the measures.m∠A = 4(20) - 10 = 70°m∠B = 6(20) - 5 = 115°m∠C = 10 + 5(20) = 110°m∠B + m∠D = 180 115 + m∠D = 180 m∠D = 65°


12. m∠MNL = 1 _ 2 m � ML

43 = 1 _ 2 m � ML

m � ML = 86°

13. m∠KMN = 1 _ 2 m � KN

= 1 _ 2 (95)

= 47.5°

14. m∠EJH = 1 _ 2 m � EGH

139 = 1 _ 2 m � EGH

m � EGH = 278°

15. m∠GFH = 1 _ 2 m � GH

= 1 _ 2 (95.2)

= 47.6° 16. m∠ADC = m∠ADB + m∠BDC

= 1 _ 2 mAB + m∠BEC

= 1 _ 2 (44) + 40 = 62°

17. m∠SRT = 903 y 2 - 18 = 90 3 y 2 = 108 y 2 = 36 y = ±6

18. �JKL is a 30°-60°-90°6z - 4 = 60 6z = 64 z = 64 __

6 = 10 2 _

3

19. m∠ADB = m∠ACB 2 x 2 = 10x 2x = 10 (x ≠ 0) x = 5m∠ADB = 1 _

2 m � AB

2(5 ) 2 = 1 _ 2 m � AB

50 = 1 _ 2 m � AB

m � AB = 2(50) = 100° 20. ∠MPN and ∠MNP are complementary.

m∠MPN + m∠MNP = 90 11x ___

3 + 3x - 10 = 90

20x - 30 = 270 20x = 300 x = 15

m∠MPN = 11(15)

______ 3 = 55°

21. Step 1 Find the value of x.m∠B + m∠D = 180 x __

2 + x __

4 + 30 = 180

3x ___ 4 = 150

3x = 600 x = 200Step 2 Find the measures.m∠B = 200 ___

2 = 100°

m∠D = 200 ___ 4 + 30 = 80°

m∠E = 200 - 59 = 141°m∠C + m∠E = 180 m∠C + 141 = 180 m∠C = 39°

22. Step 1 Find the value of x. m∠U + m∠W = 18014 + 4x + 6x - 14 = 180 10x = 180 x = 18Step 2 Find the value of y. m∠T + m∠V = 18012y - 5 + 15y - 4 = 180 27y = 189 y = 7Step 3 Find the measures.m∠T = 12(7) - 5 = 79°m∠U = 14 + 4(18) = 86°m∠V = 15(7) - 4 = 101°m∠W = 6(18) - 14 = 94°

23. always 24. never

25. sometimes (if opposite are supplementary)

26. m∠ABC = 1 _ 2 m � AC = 1 _

2 m∠ADC = 1 _

2 (112) = 56°

27. Let S be any point on the major arc from P to R. m∠PQR + m∠PSR = 180m∠PQR + 1 _

2 m � PQR = 180

m∠PQR + 1 _ 2 (130) = 180

m∠PQR = 180 - 65 = 115°

28. By the definition of an arc measure, m � JK = m∠JHK. Also, the measure of an ∠ inscribed in a circle is half the measure of the intercepted arc. So m∠JLK = 1 _

2 m � JK . Multiplying both sides of equation

by 2 gives 2m∠JLK = m � JK . By substitution, m∠JHK = 2m∠JLK.

29a. m∠BAC = 1 _ 2 m � BC = 1 _

2 ( 1 _

6 (360)) = 30°

b. m∠CDE = 1 _ 2 mCAE = 1 _

2 ( 4 _

6 (360)) = 120°

c. ∠FBC is inscribed in a semicircle. So it must be a right ∠; therefore, �FBC is a right ∠. (Also, m∠CFD = 30°. So, �FBC is a 30°-60°-90° �.)


30.

Since any 2 points determine a line, draw �� BX .

Let D be a point where �� BX intercepts � AC . By Case 1

of the Inscribed ∠ Thm., m∠ABD = 1 _ 2 m � AD and

m∠DBC = 1 _ 2 m � DC . By Add., Distrib., and Trans.

Props. of =, m∠ABD + m∠DBC = 1 _ 2 m � AD + 1 _

2 m � DC

= 1 _ 2 (m � AD + m � DC ). Thus, by ∠ Add. Post. and Arc

Add. Post., m∠ABC = 1 _ 2 mAC.

31.

Since any 2 points determine a line, draw �� BX .

Let D be pt. where �� BX intercepts � ACB . By Case 1

of Inscribed ∠ Thm., m∠ABD = 1 _ 2 m � AD and m∠CBD

= 1 _ 2 m � CD . By Subtr., Distrib., and Trans. Props.

of =, m∠ABD - m∠CBD = 1 _ 2 m � AD - 1 _

2 m � CD

= 1 _ 2 (m � AD - m � CD ). Thus by ∠ Add. Post. and Arc

Add. Post., m∠ABC = 1 _ 2 mAC.

32. By the Inscribed ∠ Thm., m∠ACD = 1 _ 2 m � AB

and m∠ADB = 1 _ 2 m � AB . By Substitution, m∠ACD =

m∠ADB, and therefore, ∠ACD � ∠ADB.

33. m∠J = 1 _ 2 m � KLM = 1 _

2 (216) = 108°

m∠J + m∠L = 180 108 + m∠L = 180 m∠L = 72°m∠M = 1 _

2 m � JKL = 1 _

2 (198) = 99°

m∠K + m∠M = 180 m∠K + 99 = 180 m∠K = 81°

34. −− PR is a diag. of PQRS. ∠Q is an inscribed right angle. So its intercepted arc is a semicircle. Thus,

−− PR is a diameter of the circle.

35a. AB 2 + AC 2 = BC 2 , so by Conv. of Pythag. Thm., �ABC is a right � with right ∠A. Since ∠A is an inscribed right ∠, it intercepts a semicircle. This means that

−− BC is a diameter.

b. m∠ABC = sin -1 - ( 14 ___ 18

) ≈ 51.1°.

Since m∠ABC = 1 _ 2 m � AC , m � AC = 102°.

36.

Draw a diagram through D and A. Label the intersection of

−− BC and −− DE as F and the intersection

of �� DA and

−− BE as G. Since −− BC is a diameter of the

circle, it is a bisector of chord −− DE . Thus,

−− DF � −− EF ,

and ∠BFD and ∠BFE are � right . −− BF �

−− BF by Reflex. Prop. of �. Thus, �BFD � �BFE by SAS.

−− BD � −− BE by CPCTC. By Trans. Prop. of �,

−− BE �

−− ED . Thus, by definition, �DBE is equilateral.

37. Agree; the opposite of a quadrilateral are congruent. So the ∠ opposite the 30° ∠ also measures 30°. Since this pair of the opposite are not supplementary, the quadrilateral cannot be inscribed in a circle.

38. Check students’ constructions.

TEST PREP, PAGE 778

39. Dm∠BAC + m∠ABC + m∠ACB = 180 m∠BAC + 1 _

2 mAC + m∠ACB = 180

m∠BAC + 1 _ 2 (76) + 61 = 180

m∠BAC + 38 + 61 = 180 m∠BAC = 81°

40. H 1 _ 2 m � XY = m∠XCY = 1 _

2 m∠XCZ

m � XY = m∠XCZ = 60° 41. C

Let m∠A = 4x and m∠C = 5x.m∠A + m∠C = 180 4x + 5x = 180 9x = 180 x = 20m∠A = 4(20) = 80°

42. Fm∠STR = 180 - (m∠TRS + m∠TSR)

= 180 - ( 1 _ 2 m � PS + 1 _

2 m � QR )

= 180 - (42 + 56) = 82°m∠QPR = 1 _

2 m � QR = 56°

m∠QPR = 1 _ 2 m � QR = 56°

m∠PQS = 1 _ 2 m � PS = 42°


43. If an ∠ is inscribed in a semicircle, the measure of the intercepted arc is 180°. The measure of ∠ is 1 _

2 (180) = 90°. So the angle is a right ∠.

Conversely, if an inscribed angle is a right angle, then it measures 90° and its intercepted arc measures 2(90) = 180°. An arc that measures 180° is a semicircle.



44. Suppose the quadrilateral ABCD is inscribed in a

circle. Then m∠A = 1 _ 2 m � BCD and m∠C = 1 _

2 m � DAB .

By Add., Distrib., and Trans. Prop. of =, m∠A +m∠C = 1 _

2 m � BCD + 1 _

2 m � DAB = 1 _

2 (m � BCD + m � DAB ).

m � BCD + m � DAB = 360°. So by subst., m∠A + m∠C= 1 _

2 (360) = 180°. Thus, ∠A and ∠C are

supplementary. A similar proof shows that ∠B and ∠D are supplementary.

45. −− RQ is a diameter. So ∠P is a right ∠. m � PQ = 2m∠R = 2 ta n -1 ( 7 _

3 ) ≈ 134°

46. Draw −− AC and

−− DE . m 1 _

2 m

−− CE = 19°m∠ACD = 1 _

2 m � AD = 36°

m∠ABC + 19 + 36 = 180 m∠ABC = 125°m∠ABD + 125 = 180 m∠ABD = 55°

47. Check students’ constructions.


48. �

�

a + b + c = 12 (1)

30a + 22.5b + 15c = 255 (2)

30a = 15c (3)

Substitute (3) in (2):15c + 22.5b + 15c = 255 3b + 4c = 34 (4)Substitute 1 __

15 (3) in 2(1):

c + 2b + 2c = 24 2b + 3c = 24 (5)3(5) - 2(4):6b + 9c - 6b - 8c = 72 - 68 c = 4Substitute in (3):30a = 15(4) = 60 a = 2Substitute in (5):2b + 3(4) = 24 2b = 12 b = 6

49. m = 1 _ 2 - (-6)

________ 8 - 4 1 _

2 =

6 1 _ 2 ___

3 1 _ 2 = 13 ___

7

50. m = -2 - (-8)

_________ 0 - (-9)

= 6 __ 9 = 2 __

3

51. m = 6 - (-14)

_________ 11 - 3

= 20 ___ 8 = 5 __

2

52. By Vert. Thm., ∠RWV � ∠SWT. So by Thm. 11-2-2 (1), RV = ST2z + 15 = 9z + 1 14 = 7z z = 2 � RV � � ST and � RS � � VT . So by substitution, 2m � ST + 2m � VT = 360 m � ST + m � VT = 180m � ST + 31(2) + 2 = 180 m � ST + 64 = 180 m � ST = 116°

53. Let BD = 2x; then 1. 5 2 + x 2 = (1 + 1.5 ) 2 2.25 + x 2 = 6.25 x 2 = 4 x = 2.�ABD has base 2(2) = 4 m and height 1.5 m. So A = 1 _

2 bh = 1 _

2 (4)(1.5) = 3 m 2 .

CONSTRUCTION, PAGE 779

1. Yes; −− CR is a radius of circle C. If a line is tangent

to a circle, then it is ⊥ to the radius at the point of tangency.

11-5 TECHNOLOGY LAB: EXPLORE ANGLE RELATIONSHIPS IN CIRCLES, PAGES 780–781

TRY THIS, PAGE 781

1. The relationship is the same. Both types of have a measure equal to half the measure of the arc.

2. A tangent line must be ⊥ to the radius at the point of tangency. If construction is done without this ⊥ relationship, then tangent line created is not guaranteed to intersect circle at only one point. If a line is ⊥ to a radius of a circle at a point on the circle, then line is tangent to circle.

3. The relationship remains the same. The measure of ∠ is half difference of intercepted arcs.

4. An ∠ whose vertex is on the circle (inscribed and created by a tangent and a secant intersecting at the point of tangency) will have a measure equal to half its intercepted arc.An ∠ whose vertex is inside the circle ( created by intersecting secants or chords) will have a measure equal to half sum of its intercepted arcs.An ∠ whose vertex is outside the circle ( created by secants and/or tangents that intersect outside circle) will have a measure equal to half its intercepted arc.

5. No; it is a means to discover relationships and make conjectures.


11-5 ANGLE RELATIONSHIPS IN CIRCLES, PAGES 782–789


1a. m∠STU = 1 _ 2 m � ST

= 1 _ 2 (166) = 83°

b. m∠QSR = 1 _ 2 m � SR

71 = 1 _ 2 m � SR

m � SR = 142°

2a. m∠ABD = 1 _ 2 (m � AD + m � CE )

= 1 _ 2 (65 + 37)

= 1 _ 2 (102) = 51°

b. m∠RNP = 1 _ 2 (m � MQ + m � RP )

= 1 _ 2 (91 + 225)

= 1 _ 2 (316) = 158°

m∠RNM + m∠RNP = 180 m∠RNM + 158 = 180 m∠RNM = 22°

3. m∠L = 1 _ 2 (m � JN - m � KN )

25 = 1 _ 2 (83 - x)

50 = 83 - x x = 33

4. m∠ACB = 1 _ 2 (m � AEB - m � AB )

= 1 _ 2 (225 - 135) = 45°

5. Step 1 Find m � PR .m∠Q = 1 _

2 (m � MS - mPR)

26 = 1 _ 2 (80 - m � PR )

52 = 80 - m � PR m � PR = 28°Step 2 Find m � LP .m � LP + m � PR = mLR m � LP + 28 = 100 m � LP = 72°


1. For both chords and secants that intersect in the interior of a circle, the measure of ∠ formed is half the sum of the measures of their intercepted arcs.

2.


GUIDED PRACTICE, PAGES 786–787

1. m∠DAB = 1 _ 2 m � AB

= 1 _ 2 (140) = 70°

2. 27 = 1 _ 2 m � AC

m � AC = 54°

3. 61 = 1 _ 2 m � PN

m � PN = 122° 4. m∠MNP = 1 _

2 (238)

= 119°

5. m∠STU = 1 _ 2 (m � SU + m � VW )

= 1 _ 2 (104 + 30)

= 1 _ 2 (134) = 67°

6. m∠HFG = 1 _ 2 (m � EJ + m � GH )

= 1 _ 2 (59 + 23)

= 1 _ 2 (82) = 41°

7. m∠NPL = 1 _ 2 (m � KM + m � NL )

= 1 _ 2 (61 + 111)

= 1 _ 2 (172) = 86°

m∠NPK + m∠NPL = 180 m∠NPK + 86 = 180 m∠NPK = 94°

8. x = 1 _ 2 (161 - 67) = 1 _

2 (94) = 47

9. x = 1 _ 2 (238 - 122) = 1 _

2 (116) = 58

10. 27 = 1 _ 2 (x - 40)

54 = x - 40 x = 94°

11. m∠S = 1 _ 2 (m � ACB - m � AB )

38 = 1 _ 2 ((360 - x) - x)

76 = 360 - 2x 2x = 360 - 76 = 284 x = 142°

12. m∠E = 1 _ 2 (m � BF - m � DF )

50 = 1 _ 2 (150 - mDF)

100 = 150 - m � DF m � DF = 50°

13. m � BC + m � CD + m � DF + m � BF = 360 64 + m � CD + 50 + 150 = 360 m � CD + 264 = 360 m � CD = 96°

14. m∠NPQ = 1 _ 2 (m � JK + m � PN )

79 = 1 _ 2 (48 + m � PN )

158 = 48 + m � PN m � PN = 110°

15. m � KN + m � PN + m � JP + m � JL = 360 m � KN + 110 + 86 + 48 = 360 m � KN + 244 = 360 m � KN = 116°



16. m∠BCD = 1 _ 2 m � BC = 1 _

2 (112) = 56°

17. m∠ABC = 1 _ 2 (360 - 112) = 1 _

2 (248) = 124°

18. m∠XZW = 1 _ 2 m � XZ = 1 _

2 (180) = 90°

19. m � XZV = m � XZ + m � ZV = 180 + 2m∠VXZ = 180 + 2(40) = 260°

20. m∠QPR = 1 _ 2 (m � QR + m � ST ) = 1 _

2 (31 + 98) = 64.5°

21. m∠ABC = 180 - m∠ABD = 180 - 1 _

2 (m � AD + m � CE )

= 180 - 1 _ 2 (100 + 45) = 107.5°

22. m∠MKJ = 180 - m∠MKL = 180 - 1 _

2 (m � JN + m � ML )

= 180 - 1 _ 2 (38.5 + 51.5) = 135°

23. x = 1 _ 2 (170 - (360 - (170 + 135)))

= 1 _ 2 (170 - 55) = 57.5°

24. x = 1 _ 2 (220 - 140) = 40°

25. x = 1 _ 2 ((180 - (20 + 104)) - 20)

= 1 _ 2 (56 - 20) = 18°

26. m∠ABV = 1 _ 2 (180 - m � AB ) = 1 _

2 (180 - 48) = 66°

27. m∠DJH = 1 _ 2 (m � DH + m � EG )

180 - 89 = 1 _ 2 (137 + m � EG )

91 = 1 _ 2 (137 + m � EG )

182 = 137 + m � EG m � EG = 45°

28. m∠DJE = 1 _ 2 (m � DE + m � GH )

89 = 1 _ 2 (m � DE + 61)

178 = m � DE + 61 m � DE = 117°

29. Step 1 Find m∠P.m∠P = 1 _

2 m � LR = 1 _

2 (170) = 85°

Step 2 Find m∠QPR.m∠QPR + 50 + 85 = 180 m∠QPR = 45°Step 3 Find m � PR .m∠QPR = 1 _

2 m � PR

45 = 1 _ 2 m � PR

m � PR = 90°

30. 50 = 1 _ 2 (m � PS - m � ML )

= 1 _ 2 (m � PM - m � ML )

= 1 _ 2 m � LP

m � LP = 100°

31. m∠ABC = 1 _ 2 m � AB

x = 1 _ 2 m � AB

m � AB = 2x ° 32. m∠ABD + m∠ABC = 180

m∠ABD + x = 180 m∠ABD = (180 - x)°

33. m � AEB + m � AB = 360

m � AEB + 2x = 360

m � AEB = (360 - 2x)°

34.

Since 2 points determine a line, draw −− BD . By

Ext. ∠ Thm., m∠ABD = m∠C + m∠BDC. So

m∠C = m∠ABD - m∠BDC. m∠ABD = 1 _ 2 m � AD by

Inscribed ∠ Thm., and m∠BDC = 1 _ 2 m � BD by Thm.

11-5-1. By substitution, m∠C = 1 _ 2 m � AD - 1 _

2 m � BD .

Thus, by Distrib. Prop. of =, m∠C = 1 _ 2 (m � AD

- m � BD ).

35.

Since 2 points determine a line, draw −− EG . By

Ext. ∠ Thm., m∠DEG = m∠F + m∠EGF. So

m∠F = m∠DEG - m∠EGF. m∠DEG = 1 _ 2 m � EHG

and m∠EGF = 1 _ 2 m � EG by Theorem 11-5-1. By

substitution, m∠F = 1 _ 2 m � EHG - 1 _

2 m � EG . Thus, by

Distrib. Prop. of =, m∠F = 1 _ 2 (m � EHF - m � EG ).

36.

Since 2 points determine a line, draw −− JM . By

Ext. ∠ Thm., m∠JMN = m∠L + m∠KJM. So

m∠L = m∠JMN - m∠KJM. m∠JMN = 1 _ 2 m � JN

and m∠KJM = 1 _ 2 m � KM by Inscribed ∠ Thm. By

substitution, m∠F = 1 _ 2 m � JN - 1 _

2 m � KM . Thus, by

Distrib. Prop. of =, m∠F = 1 _ 2 (m � JN - m � KM ).

37. m∠1 > m∠2 because m∠1 = 1 _ 2 (m � AB + m � CD )

and m∠2 = 1 _ 2 (m � AB - m � CD ). Since m � CD > 0,

the expression for m∠1 is greater.



38. When a tangent and secant intersect on the circle, the measure of ∠ formed is half the measure of the intercepted arc.

When 2 secants intersect inside the circle, the measure of each ∠ formed is half the sum of the measures of the intercepted arcs, or 1 _

2 (90° + 90°).

When 2 secants intersect inside the circle, the measure of each ∠ formed is half the difference of the measures of the intercepted arcs, or 1 _ 2 (270° - 90°).

39. m � AC + m � AD + m � CD = 3602x - 10 + x + 160 = 360 3x = 210 x = 70m∠B = 1 _

2 (m � AC - m � AD ) = 1 _

2 (2(70) - 10 - 70) = 30°

m∠C = 1 _ 2 m � AD = 1 _

2 (70) = 35°

m∠A = 180 - (m∠B + m∠C) = 180 - (30 + 35) = 115°

40. m∠B = 1 _ 2 (m � AD - m � DE )

4x = 1 _ 2 (18x - 15 - (8x + 1))

8x = 10x - 16 16 = 2x x = 8 m∠B = 4(8) = 32°m � AD = 18(8) - 15 = 129°m∠C = 1 _

2 (m � AED - m � AD ) = 1 _

2 (231 - 129) = 51°

m∠A = 180 - (m∠B - m∠C) = 180 - (32 + 51) = 97°

41a. m∠BHC = m � BC = 1 _ 6 (360) = 60°

b. m∠EGD = 1 _ 2 (m � DE - m � DAE )

= 1 _ 2 (60 + 3(60)) = 120°

c. m∠CED = m∠EDF = 1 _ 2 (60) = 30° and

m∠EGD > 90°, so �EGD is an obtuse isosceles.

TEST PREP, PAGE 789

42. Cm∠DCE = 1 _

2 (m � AF + m � DE )

= 1 _ 2 (58 + 100) = 79°

43. J

44. 56°m∠JLK = 1 _

2 (m � MN - m � JK )

45 = 1 _ 2 (146 - m � JK )

90 = 146 - m � JK m � JK = 56°


45. Case 1: Assume −− AB is a diameter of the circle.

Then m � AB = 180° and ∠ABC is a right ∠. Thus, m∠ABC = 1 _

2 m � AB .

Case 2: Assume −− AB is not a diameter of the circle.

Let X be the center of the circle and draw radii −− XA

and −− XB .

−− XA � −− XB . So �AXB is isosceles. Thus,

∠XAB � ∠XBA, and 2m∠XBA + m∠AXB = 180. This means that m∠XBA = 90 - 1 _

2 m∠AXB. By

Thm. 11-1-1, ∠XBC is a right ∠. So m∠XBA + m∠ABC = 90 or m∠ABC = 90 - m∠XBA. By substituting, m∠ABC = 90 - (90 - 1 _

2 m∠AXB)

= 1 _ 2 m∠AXB. m∠AXB = m � AB because ∠AXB is

a central ∠. Thus, m∠ABC = 1 _ 2 m � AB .

46. Since m � WY = 90°, m∠YXW = 90° because it is a central ∠. By Thm. 11-1-1, ∠XYZ and ∠XWZ are right �. The sum of measures of the � of a quadrilateral is 360°. So m∠WZY = 90°. Thus, all 4 � of WXYZ are right �. So WXYZ is a rectangle. XY � XW because they are radii. By Thm. 6-5-3, WXYZ is a rhombus. Since WXYZ is a rectangle and a rhombus, it must also be a square by Theorem 6-5-6.

47. m∠V = 1 _ 2 (x - 21) = 1 _

2 (124 - 50)

x - 21 = 124 - 50 = 74 x = 74 + 21 = 95°

48. Step 1 Find m∠CED.m∠DCE + m∠ECJ = 180 m∠DCE + 135 = 180 m∠DCE = 45°m∠CDE = m∠FDG = 82°m∠CED + m∠DCE + m∠CDE = 180 m∠CED + 45 + 82 = 180 m∠CED = 53°Step 2 Find m � GH .m∠CED = 1 _

2 (m � GH + mKL)

53 = 1 _ 2 (m � GH + 27)

106 = m � GH + 27 m � GH = 79°


49. g(7) = 2(7 ) 2 - 15(7) - 1 = 98 - 105 - 1 = -8; yes

50. f(7) = 29 - 3(7) = 29 - 21 = 8; no

51. - 7 __ 8 (7) = - 49 ___

8 ; no 52. V = 1 _

3 Bh

= 1 _ 3 ( 1 _

2 aP) h

= 1 _ 3 ( 1 _

2 (2 √ 3 ) (24)) (7)

= 56 √ 3 ≈ 97.0 m 3



53. L = πr�60π = π(6)� � = 10 cm r 2 + h 2 = � 2 6 2 + h 2 = 10 2 h = 8 cmV = 1 _

3 π r 2 h

= 1 _ 3 π(6 ) 2 (8) = 96π cm 3

≈ 310.6 cm 3

54. S = 1 _ 2 P� + s 2

1200 = 1 _ 2 (96)� + 576

624 = 48�

� = 13 in. ( 1 _

2 s)

2 + h 2 = � 2

12 2 + h 2 = 13 2 h = 5 in.V = 1 _

3 Bh = 1 _

3 (576)(5) = 960 in. 3

55. m∠BCA = 1 _ 2 m � BA = 1 _

2 (74) = 37°

56. m∠DCB = 1 _ 2 m � DB = m∠DAB = 67°

m∠BDC = 90° ( � BC is a diam.)m∠DBC = 180 - (90 + 67) = 23°

57. m∠ADC = 1 _ 2 m � AC = 1 _

2 (180 - 74) = 53°

11-6 TECHNOLOGY LAB: EXPLORE SEGMENT RELATIONSHIPS IN CIRCLES, PAGES 790–791

ACTIVITY 1, TRY THIS, PAGE 790

1.

2. ∠CGF and ∠EGD; ∠FCG and ∠DEG; �CFG and �EDG are ∼ � by AA ∼ Post.

3. GC ___ GF

= GE ___ GD

; GC · GD = GE · GF


4. The products of the segment lengths for a tangent and a secant are similar to the products of the segment lengths for 2 secants because for a tangent there is only 1 segment. Thus, the “whole segment” multiplied by the “external segment” becomes the square of the tangent seg.

5. 2 tangent segments from same external point will have = lengths, so the circle segments are �.

6. In diagram, the circle A is given with tangents −− DC

and −− DE from D. Since 2 points determine a line,

draw radii −− AC ,

−− AE , and −− AD .

−− AC � −− AE because all

radii are �. −− AD �

−− AD by Reflex. Prop. of �. ∠ACD and ∠AED are right because they are each formed by a radius and a tangent intersecting at point of tangency. Thus, �ACD and �AED are right �. �ACD � �AED by HL. Therefore,

−− DC �

−− DE by CPCTC.


7. ∠DGE and ∠FGC; ∠GDE and ∠GFC; ∠GED and ∠GCF; �DGE and �FGC are ∼ � by AA ∼.

8. When 2 secants or 2 chords of a circle intersect, 4 segments will be formed, each with the point of intersection as 1 endpoint. The product of segment lengths on 1 secant/chord will equal the product of segment lengths on the other secant/chord.If a secant and a tangent of a circle intersect on an exterior point, 3 segments will be formed, each with exterior point as an endpoint. The product of the secant segment lengths will equal square of tangent segment length.

11-6 SEGMENT RELATIONSHIPS IN CIRCLES, PAGES 792–798


1. AE · EB = CE · ED 6(5) = x(8) 30 = 8x x = 3.75AB = 6 + 5 = 11CD = (3.75) + 8 = 11.75

2. original disk: PR = 11 1 _ 3 in.

New disk:AQ · QB = PQ · QR 6(6) = 3(QR) 36 = 3QR QR = 12 in.PR = 12 + 3 = 15 in.change in PR = 15 - 11 1 _

3 = 3 2 _

3 in.

3. GH · GJ = GK · GL13(13 + z) = 9(9 + 30) 169 + 13z = 81 + 270 13z = 182 z = 14GJ = 13 + (14) = 27GL = 9 + 30 = 39

4. DE · DF = DG 2 7(7 + y) = 10 2 49 + 7y = 100 7y = 51 y = 7 2 _

7



1. Yes; in this case, chords intersect at center of the circle. So segments of the chords are all radii, and theorem simplifies to r 2 = r 2 .

2. 2

3.


GUIDED PRACTICE, PAGES 795–796

1. tangent segment

2. HK · KJ = LK · KM 8(3) = 4(y) 24 = 4y y = 6LM = 4 + (6) = 10HJ = 8 + 3 = 11

3. AE · EB = CE · ED x(4) = 6(6) 4x = 36 x = 9AB = (9) + 4 = 13CD = 6 + 6 = 12

4. PS · SQ = RS · ST z(10) = 6(8) 10z = 48 z = 4.8PQ = 10 + (4.8) = 14.8RT = 6 + 8 = 14

5. Let the diameter be d ft.GF · FH = EF · (d - EF) 25(25) = 20(d - 20) 625 = 20d - 400 225 = 20d d = 51 1 _

4 ft

6. AC · BC = EC · DC(x + 7.2)7.2 = (9 + 7.2)7.2 x + 7.2 = 9 + 7.2 x = 9AC = (9) + 7.2 = 16.2EC = 9 + 7.2 = 16.2

7. PQ · PR = PS · PT 5(5 + y) = 6(6 + 7) 25 + 5y = 78 5y = 53 y = 10.6PR = 5 + (10.6) = 15.6PT = 6 + 7 = 13

8. GH · GJ = GK · GL10(10 + 10.7) = 11.5(11.5 + x) 207 = 132.25 + 11.5x 74.75 = 11.5x x = 6.5GJ = 10 + 10.7 = 20.7GL = 11.5 + (6.5) = 18

9. AB · AC = AD 2 2(2 + 6) = x 2 16 = x 2 x = 4 (since x > 0)

10. MP · MQ = MN 2 3(3 + y) = 4 2 9 + 3y = 16 3y = 7 y = 2 1 _

3

11. RT · RU = RS 2 3(3 + 8) = x 2 33 = x 2 x = √ �� 33 (since x > 0)


12. DH · HE = FH · HG 3(4) = 2(y) 12 = 2y y = 6DE = 3 + 4 = 7FG = 2 + (6) = 8

13. JK · KL = MN · KN 10(x) = 6(7) 10x = 42 x = 4.2JL = 10 + (4.2) = 14.2MN = 6 + 7 = 13

14. UY · YV = WY · YZ 8(5) = x(11) 40 = 11x x = 3 7 __

11

UV = 8 +5 = 13WZ = 3 7 __

11 + 11 = 14 7 __

11

15. 590(590) = 225.4(d - 225.4) 348,100 = 225.4d - 50,805.16398,905.16 = 225.4d d ≈ 1770 ft

16. AB · AC = AD · AE5(5 + x) = 6(6 + 6) 25 + 5x = 72 5x = 47 x = 9.4AC = 5 + (9.4) = 14.4AE = 6 + 6 = 12

17. HL · JL = NL · ML(y + 10)10 = (18 + 9)9 10y + 100 = 243 10y = 143 y = 14.3HL = (14.3) + 10 = 24.3NL = 18 + 9 = 27

18. PQ · PR = PS · PT 3(3 + 6) = 2(2 + x) 27 = 4 + 2x 23 = 2x x = 11.5PR = 3 + 6 = 9PT = 2 + (11.5) = 13.5

19. UW · UX = UV 2 6(6 + 8) = z 2 84 = z 2 z = √ �� 84 = 2 √ �� 21 (since z > 0)


20. BC · BD = BA 2 2(2 + x) = 5 2 4 + 2x = 25 2x = 21 x = 10.5

21. GE · GF = GH 2 8(8 + 12) = y 2 160 = y 2 y = 4 √ �� 10 (since y > 0)

22a. RM · MS = PM · MQ 10(MS) = 12(12) = 144 MS = 14.4 cm

b. RS = RM + MS = 10 + 14.4 = 24.4 cm

23a. PM · MQ = RM · MS PM 2 = 4(13 - 4) = 36 PM = 6 in.

b. PQ = 2PM = 2(6) = 12 in.

24. Step 1 Find x. AB · AC = AD · AE5(5 + 5.4) = 4(4 + x) 52 = 16 + 4x 36 = 4x x = 9Step 2 Find y.AB · AC = AF 2 52 = y 2 y = √ �� 52 = 2 √ �� 13

25. Step 1 Find x.NQ · QM = PQ · QR 4(x) = 3.2(10) = 32 x = 8Step 2 Find y. KM · KN = KL 2 6(6 + (8) + 4) = y 2 108 = y 2 y = √ �� 108 = 6 √ � 3

26. SP 2 = SE · (SE + d) = 6000(14,000) = 84,000,000 SP = √ �� 84,000,000 = 2000 √ �� 21 ≈ 9165 mi

27. Solution B is incorrect. The first step should be AC · BC = DC 2 , not AB · BC = DC 2 .

28.

Since 2 points determine a line, draw −− AC and

−− BD . ∠ACD � ∠DBA because they both intercept � AD . ∠CEA � ∠BED by Vert. Thm. Therefore, �ECA ∼ �EBD by AA ∼. Corresponding sides are proportional. So AE ___

ED = CE ___

EB . By Cross Products

Property, AE · EB = CE · ED.

29.

Since 2 points determine a line, draw −− AD and

−− BD . m∠CAD = 1 _

2 m � BD by Inscribed ∠ Thm. m∠BDC =

1 _ 2 m � BD by Thm. 11-5-1. Thus, ∠CAD � ∠BDC. Also,

∠C � ∠C by Reflex. Prop. of �. Therefore, �CAD ∼ �CDB by AA ∼. Corresponding sides are proportional. So AC ___

DC = DC ___

BC . By Cross Products

Property, AC · BC = DC 2 .

30. Yes; PR · PQ = PT · PS, and it is given that PQ = PS.So PR = PT. Subtracting the � segments from each of these shows that

−− QR � −− ST .

31. Method 1: By Secant-Tangent Product Theorem, BC 2 = 12(4) = 48, and so BC =

√ �� 48 = 4

√ � 3 .

Method 2: By Thm. 11-1-1, ∠ABC is a right ∠. By Pythagorean Thm., BC 2 + 4 2 = 8 2 . Thus BC 2 = 64 - 16 = 48, and BC = 4

√ � 3 .

32a. AE · EC = BE · DE 5.2(4) = 3(DE) 20.8 = 3DE DE ≈ 6.9 cm

b. diameter = BD = BE + DE ≈ 3 + 6.9 ≈ 9.9 cm

c. OE = OB - BE ≈ 1 _ 2 (9.933) - 3 ≈ 1.97 cm

TEST PREP, PAGE 798

33. B PQ 2 = PR · PS = 6(6 + 8) = 84 PQ = √ �� 84 ≈ 9.2

34. F PT · PU = PQ 2 7(7 + UT) = 84 7 + UT = 12 UT = 5

35. CE = ED = 6, and by Chord-Chord Product Thm., 3(EF) = 6(6) = 36. So EF = 12, FB = 12 + 3 = 15, and radius AB = 1 _

2 (15) = 7.5.


36a. Step 1 Find the length y of the chord formed by

−− KM . KL 2 = 144 = 8(8 + y)

18 = 8 + y y = 10

Step 2 Find the value of x.6(6 + x) = 8(8 + (10)) 36 + 6x = 144 6x = 108 x = 18

b. KL 2 + LM 2 � KM 2 12 2 + (6 + 18 ) 2 � (8 + 10 + 8 ) 2 720 > 676�KLM is acute.


37. Let R be center of the circle; then �PQR is a right ∠. So

PR 2 = PQ 2 + QR 2 = 6 2 + 4 2 = 52 PR = √ �� 52 = 2 √ �� 13 in.Let S be the intersection of

−− PR with circle R, so that SR = 4 in.; then the distance from P to the circle = PS = 2 √ �� 13 - 4 ≈ 3.2 in.

38. By Chord-Chord Product Thm., (c + a)(c - a) = b · b c 2 - a 2 = b 2 c 2 = a 2 + b 2

39. GJ · HJ = FJ 2 (y + 6)y = 10 2 = 100 y 2 + 6y - 100 = 0

y = -6 ± √ �� 6 2 - 4(1)(-100)

____________________ 2

= -6 + √ �� 436

__________ 2 (since y > 0)

= -3 + √ �� 109 ≈ 7.44


40. P = # favorable outcomes __________________ # trials

0.035 = 14 ___ n

n = 14 _____ 0.035

= 400

41. P = 36 ___ 50

= 0.72 = 72%

42. �� BA and

�� CD 43.

�� CD 44.

−− BC

45. A = π(12 ) 2 ( 55 ____ 360

) = 22π ft 2 ≈ 69.12 ft 2

46. L = 2π(12) ( 55 ____ 360

) = 3 2 __ 3 π ft ≈ 11.52 ft

47. The area of sector YZX is 40π - 22π = 18π ft 2

A = π r 2 ( m ____ 360

)

18π = π(12 ) 2 ( m ____ 360

)

m ____ 360

= 18 ____ 144

= 1 __ 8

m = 1 __ 8 (360) = 45°

11-7 CIRCLES IN THE COORDINATE PLANE, PAGES 799–805


1a. (x - h ) 2 + (y - k ) 2 = r 2 (x - 0 ) 2 + (y - (-3) ) 2 = 8 2 x 2 + (y + 3 ) 2 = 64

b. r = √ �� (2 - 2 ) 2 + (3 - (-1) ) 2 = √ �� 16 = 4(x - 2 ) 2 + (y - (-1) ) 2 = 4 2 (x - 2 ) 2 + (y + 1 ) 2 = 16

2a. Step 1 Make a table of values.Since the radius is √ � 9 = 3, use ±3 and values in between for x-values.

x -3 -2 -1 0 -1 -2 -3

y 0 ±2.2 ±2.8 ±3 ±2.8 ±2.2 0

Step 2 Plot points and connect them to form a circle.

b. The equation of given circle can be written as (x - 3 ) 2 + (y - (-2) ) 2 = 2 2 So h = 3, k = -2, and r = 2. The center is (3, -2) and the radius is 2. Plot point (3, -2). Then graph a circle having this center and radius 2.

3. Step 1 Plot 3 given points.Step 2 Connect D, E, and F to form a �.

Step 3 Find a point that is equidistant from 3 points by constructing ⊥ bisectors of 2 sides of �DEF. The ⊥ bisectors of sides of �DEF intersect at a point that is equidistant from D, E, and F. The intersection of the ⊥ bisectors is P(2, -1). P is the center of circle passing through D, E, and F.



1. x 2 + y 2 = r 2

2. First find the center by finding the midpoint of the diameter. By the Midpoint Formula, the center of the circle is (-1, 4). The radius is half the length of the diameter. So r = 2. The equation is (x + 1 ) 2 + (y - 4 ) 2 = 4.

3. No; a radius represents length, and length cannot be negative.

4.



1. (x - h ) 2 + (y - k ) 2 = r 2 (x - 3 ) 2 + (y - (-5) ) 2 = 12 2 (x - 3 ) 2 + (y + 5 ) 2 = 144

2. (x - h ) 2 + (y - k ) 2 = r 2 (x - (-4) ) 2 + (y - 0 ) 2 = 7 2 (x + 4 ) 2 + y 2 = 49

3. r = √ �� (2 - 4 ) 2 + (0 - 0 ) 2 = √ � 4 = 2(x - 2 ) 2 + (y - 0 ) 2 = 2 2 (x - 2 ) 2 + y 2 = 4

4. r = √ �� (2 - (-1) ) 2 + (-2 - 2 ) 2 = √ �� 25 = 5(x - (-1) ) 2 + (y - 2 ) 2 = 5 2 (x + 1 ) 2 + (y - 2 ) 2 = 25

5. The equation of given circle can be written as (x - 3 ) 2 + (y - 3 ) 2 = 2 2 So h = 3, k = 3, and r = 2. The center is (3, 3) and radius is 2. Plot point (3, 3). Then graph a circle having this center and radius 2.

6. The equation of given circle can be written as (x - 1 ) 2 + (y - (-2) ) 2 = 3 2 So h = 1, k = -2, and r = 3. The center is (1, -2) and radius is 3. Plot point (1, -2). Then graph a circle having this center and radius 3.

7. The equation of given circle can be written as (x - (-3) ) 2 + (y - (-4) ) 2 = 1 2 So h = -3, k = -4, and r = 1. The center is (-3, -4) and radius is 1. Plot point (-3, -4). Then graph a circle having this center and radius 1.

8. The equation of given circle can be written as (x - 3 ) 2 + (y - (-4) ) 2 = 2 2 So h = 3, k = -4, and r = 2. The center is (3, -4) and radius is 4. Plot point (3, -4). Then graph a circle having this center and radius 4.

9a. Step 1 Plot the 3 given points.Step 2 Connect A, B, and C to form a �.

Step 3 Find a point that is equidistant from 3 points by constructing ⊥ bisectors of 2 sides of �ABC.The ⊥ bisectors of sides of �ABC intersect at a point that is equidistant from A, B, and C.The intersection of ⊥ bisectors is P(-2, 3). P is center of circle passing through A, B, and C.

b. There are approximately 10 units across the circle. So the diameter is approximately 10 ft.




10. (x - h ) 2 + (y - k ) 2 = r 2 (x - (-12) ) 2 + (y - (-10) ) 2 = 8 2 (x + 12 ) 2 + (y + 10 ) 2 = 64

11. (x - h ) 2 + (y - k ) 2 = r 2 (x - 1.5 ) 2 + (y - (-2.5) ) 2 = ( √ � 3 )

2

(x - 1.5 ) 2 + (y + 2.5 ) 2 = 3

12. r = √ �� (2 - 1 ) 2 + (2 - 1 ) 2 = √ � 2 (x - 1 ) 2 + (y - 1 ) 2 = ( √ � 2 )

2

(x - 1 ) 2 + (y - 1 ) 2 = 2

13. r = √ �� (-5 - 1 ) 2 + (1 - (-2) ) 2 = √ �� 45 = 3 √ � 5 (x - 1 ) 2 + (y - (-2) ) 2 = (3 √ � 5 )

2

(x - 1 ) 2 + (y + 2 ) 2 = 45

14. The equation of given circle can be written as (x - 0 ) 2 + (y - 2 ) 2 = 3 2 .So h = 0, k = 2, and r = 3. The center is (0, 2)and radius is 3. Plot point (0, 2). Then graph a circle having this center and radius 3.

15. The equation of given circle can be written as (x - (-1) ) 2 + (y - 0 ) 2 = 4 2 .So h = -1, k = 0, and r = 4. The center is (-1, 0)and radius is 4. Plot point (-1, 0). Then graph a circle having this center and radius 4.

16. The equation of given circle can be written as (x - 0 ) 2 + (y - 0 ) 2 = 10 2 .So h = 0, k = 0, and r = 10. The center is (0, 0) and radius is 10. Then graph a circle having origin as center and radius 10.

17. The equation of given circle can be written as (x - 0 ) 2 + (y - (-2) ) 2 = 2 2 .So h = 0, k = -2, and r = 2. The center is (0, -2) and radius is 2. Plot point (0, -2). Then graph a circle having this center and radius 2.

18a. Step 1 Plot the 3 given points.Step 2 Connect A, B, and C to form a �.

Step 3 Find a point that is equidistant from the 3 points by constructing ⊥ bisectors of the 2 sides of �ABC.

The ⊥ bisectors of sides of �ABC intersect at a point that is equidistant from A, B, and C.

The intersection of ⊥ bisectors is P(-1, -2). P is center of the circle passing through A, B, and C.

b. There are approximately 10 units across the circle. So the diameter is approximately 10 ft.

19. The circle has the center (1, -2) and radius 2.(x - 1 ) 2 + (y - (-2) ) 2 = 2 2 (x - 1 ) 2 + (y + 2 ) 2 = 4

20. The circle has the center (-1, 1) and radius 4.(x - (-1) ) 2 + (y - 1 ) 2 = 4 2 (x + 1 ) 2 + (y - 1 ) 2 = 16

21a. r = √ �� 24 2 + 32 2 = 40d = 2(40) = 80 units or 80 ft

b. x 2 + y 2 = 40 2 x 2 + y 2 = 1600

22. F; r = √ � 7

23. T; (-1 - 2 ) 2 + (-3 + 3 ) 2 = 9

24. F; center is (6, -4), in fourth quadrant.

25. T; (0, 4 ± √ � 3 ) lie on y-axis and �.

26. F; the equation is x 2 + y 2 = 3 2 = 9.


27a. Possible answer: 28 units 2

b. r = 3, so A = π(3 ) 2 = 9π ≈ 28.3 units 2

c. Check students’ work.

28. slope of radius from center (4, -6) to pt. (1, -10) is

m = -10 - (-6)

__________ 1 - 4

= -4 ___ -3 = 4 __

3

tangent has slope - 3 __ 4 , and eqn.

y - (-10) = - 3 __ 4 (x - 1) or y + 10 = - 3 __

4 (x - 1)

29a. r = 3E(-3, 5 - 2(3)) = E(-3, -1)G(0 - 2(3), 2) = G(-6, 2)

b. d = 2(3) = 6

c. center is (0 - 3, 2) = (-3, 2) (x - (-3) ) 2 + (y - 2 ) 2 = 3 2

(x + 3 ) 2 + (y - 2 ) 2 = 9

30. (x - 2 ) 2 + (x + 3 ) 2 = 81(x - 2 ) 2 + (x - (-3) ) 2 = 9 2 center (2, -3), radius 9

31. x 2 + (y + 15 ) 2 = 25(x - 0 ) 2 + (y - (-15) ) 2 = 5 2 center (0, -15), radius 5

32. (x + 1 ) 2 + y 2 = 7(x - (-1) ) 2 + (y - 0 ) 2 = ( √ � 7 )

2

center (-1, 0), radius √ � 7

33. r = 3; A = π(3 ) 2 = 9π; C = 2π(3) = 6π

34. r = √ � 7 ; A = π ( √ � 7 ) 2 = 9π; C = 2π ( √ � 7 ) = 2 √ � 7 π

35. r = √ �� (2 - (-1) ) 2 + (-1 - 3 ) 2 = 5A = π(5 ) 2 = 25π; C = 2π(5) = 10π

36. Graph is a single point, (0, 0).

37. The epicenter (x, y) is a solution of the equations of 3 circles. Let 1 unit represent 100 mi.seismograph A: (x + 2 ) 2 + (y - 2 ) 2 = 3 2 x 2 + 4x + 4 + y 2 - 4y + 4 = 9 x 2 + 4x + y 2 - 4y = 1 (1)seismograph B: (x - 4 ) 2 + (y + 1 ) 2 = 6 2 x 2 - 8x + 16 + y 2 + 2y + 1 = 36 x 2 - 8x + y 2 + 2y = 19 (2)seismograph C: (x - 1 ) 2 + (y + 5 ) 2 = 5 2 x 2 - 2x + 1 + y 2 + 10y + 25 = 25 x 2 - 2x + y 2 + 10y = -1 (3)(1) - (2):12x - 6y = -18 2x + 3 = y (4)(1) - (3):6x - 14y = 2 3x = 7y + 1 (5)(4) in (5): 3x = 7(2x + 3) + 1 3x = 14x + 22-11x = 22 x = -2y = 2(-2) + 3 = -1The location of epicenter is (-200, -100).

38. The circle has a radius of 5. So 5 is tangent to x-axis if the center has y-coordinate k = ±5.

39. d = √ �� (5 - (-3) ) 2 + (-2 - (-2) ) 2 = 8; r = 4

center = ( -3 + 5 _______ 2 ,

-2 + (-2) _________

2 ) = (1, -2)

equation is (x - 1 ) 2 + (y + 2 ) 2 = 16

40. The locus is a circle with a radius 3 centered at (2, 2).

41. The point does not lie on circle P because it is not a solution to the equation (x - 2 ) 2 + (y - 1 ) 2 = 9. Since (3 - 2 ) 2 + ((-1) - 1 ) 2 < 9, the point lies in the interior of circle P.

TEST PREP, PAGE 804

42. C(-2, 0) lies on the circle.

43. H(x - (-3) ) 2 + (y - 5 ) 2 = (1 - (-3) ) 2 + (5 - 5 ) 2 (x + 3 ) 2 + (y - 5 ) 2 = 16

44. Adistances from statues to fountain:

√ �� (4 - (-1) ) 2 + (-2 - (-2) ) 2 = 5

√ �� (-1 - (-1) ) 2 + (3 - (-2) ) 2 = 5

√ �� (-5 - (-1) ) 2 + (-5 - (-2) ) 2 = 5



45a. r = √ �� (1 - 2 ) 2 + (-2 - (-4) ) 2 + (-5 - 3 ) 2 = √ �� 69

(x - 2 ) 2 + (y - (-4) ) 2 + (z - 3 ) 2 = ( √ �� 69 ) 2

(x - 2 ) 2 + (y + 4 ) 2 + (z - 3 ) 2 = 69

b. 15; if 2 segments are tangent to a circle or sphere from same exterior point, then segments are �.

46. x + y = 5 y = 5 - xSubstitute in equation of a circle: x 2 + (5 - x ) 2 = 25 x 2 + 25 - 10x + x 2 = 25 2 x 2 - 10x = 0 2x(x - 5) = 0 x = 0 or 5The point of intersection are (0, 5 - (0)) = (0, 5) and (5, 5 - (5)) = (5, 0).

47. Given the line is ⊥ to a line through (3, 4) with slope -0.5. For point of tangency,y = 2x + 3 (1)and y - 4 = -0.5(x - 3)2(y - 4) = 3 - x 2y - 8 = 3 - x x = 11 - 2y (2)(2) in (1): y = 2(11 - 2y) + 3 y = 25 - 4y5y = 25 y = 5x = 11 - 2(5) = 1Point of tangency is (1, 5). So r 2 = (1 - 3 ) 2 + (5 - 4 ) 2 = 5.The equation of the circle is(x - 3 ) 2 + (y - 4 ) 2 = 5


48. 2 x 2 - 2(4 x 2 + 1)

_______________ 2

x 2 - (4 x 2 + 1) x 2 - 4 x 2 - 1-3 x 2 - 1

49. 18a + 4(9a + 3)

______________ 6

3a + 2 _ 3 (9a + 3)

3a + 6a + 29a + 2

50. 3(x + 3y) - 4(3x + 2y) - (x - 2y)3x + 9y - 12x - 8y - x + 2y-10x + 3y

51. By the Isosc. � Thm., m∠D = m∠F7x + 4 = 60 7x = 56 x = 8

52. DE = EF2y + 10 = 4y - 1 11 = 2y y = 5.5

53. 180 - 142 = 38 = 1 _ 2 (m � LK + m � JQ )

m � LKQ = m � LK + m � KJ + m � JQ = 2(38) + 88 = 164°m � LNQ = 360 - 164 = 196°

54. m∠NMP = 1 _ 2 (m � LKQ - m � NP )

= 1 _ 2 (164 - 50) = 57°

11B MULTI-STEP TEST PREP, PAGE 806

1. m∠AGB = 1 _ 2 m � AB = 1 _

2 ( 1 __

12 (360)) = 15°

2. m∠KAE = 90°; the angle is inscribed in a semicircle. So it is a right ∠.

3. m∠KMJ = 1 _ 2 (m � KJ + m � BE ) = 1 _

2 (30 + 90) = 60°

4. ME = 22 - 4.8 ME ≈ 17.2 KM · ME = JM · MB

4.8 · 17.2 = 6.4x 82.56 = 6.4x 12.9 ≈ x MB ≈ 12.9 ft

5. (x - 20 ) 2 + (y - 14 ) 2 = 11 2 = 121

6. L(20, 14 + 11) = L(20, 25)C(20 + 11, 14) = C(31, 14)F(20, 14 - 11) = F(20, 3)I(20 - 11, 14) = I(9, 14)

11B READY TO GO ON? PAGE 807

1. m∠BAC = 1 _ 2 m � BC = 1 _

2 (102) = 51°

2. m � CD = 2m∠CAD = 2(38) = 76° 3. ∠FGH is inscribed in a semicircle.

So m∠FGH = 90°. 4. m � JGF = 310° 5. m∠RST = 1 _

2 mST = 1 _

2 (266) = 133°

6. m∠AEC = 1 _ 2 (m � AC + m � BD ) = 1 _

2 (130 + 22) = 76°

7. m∠MPN = 1 _ 2 ((360 - 102) - 102) = 78°

8. AE · EB = CE · ED 2(6) = x(3) 12 = 3x x = 4AB = 2 + 6 = 8CD = (4) + 3 = 7

9. FH · GH = KH · JH (y + 3)3 = (3 + 4)4 3y + 9 = 28 3y = 19 y = 6 1 _

3

FH = (6 1 _ 3 ) + 3 = 9 1 _

3

KH = 3 + 4 = 7

10. RU · (d - UR) = SU · UT 3.9(d - 3.9) = 6.1(6.1) 3.9d - 15.21 = 37.21 3.9d = 52.42 d ≈ 13.44 m

11. (x - (-2) ) 2 + (y - (-3) ) 2 = 3 2 (x + 2 ) 2 + (y + 3 ) 2 = 9

12. r = √ �� (1 - 4 ) 2 + (1 - 5 ) 2 = 5(x - 4 ) 2 + (y - 5 ) 2 = 5 2 (x - 4 ) 2 + (y - 5 ) 2 = 25


ge07_SOLKEY_C11_273-300.indd 296ge07_SOLKEY_C11_273-300.indd 296 10/5/06 11:44:51 AM10/5/06 11:44:51 AM

13. Step 1 Plot the 3 given points.Step 2 Connect J, K, and L to form a �.

Step 3 Find a point that is equidistant from the 3 points by constructing ⊥ bisectors of 2 sides of �JKL. The ⊥ bisectors of the sides of �JKL intersect at a point that is equidistant from A, B, and C. The intersection of the ⊥ bisectors is P(-1, -2). P is the center of the circle passing through J, K, and L.

STUDY GUIDE: REVIEW, PAGES 810–813

VOCABULARY, PAGE 810 1. segment of a circle 2. central angle

3. major arc 4. concentric circles

LESSON 11-1, PAGE 810 5. chords:

−− QS , −− UV ; tangent: �; radii:

−− PQ , −− PS ; secant: � �� UV ;

diameter: −− QS

6. chords: −− KH ,

−−− MN ; tangent: � �� KL ; radii: −− JH ,

−− JK , −− JM ,

−− JN ; secant: � �� MN ; diameters:

−− KH , −−− MN

7. AB = BC9x - 2 = 7x + 4 2x = 6 x = 3AB = 9(3) - 2 = 25

8. EF = EG5y + 32 = 8 - y 6y = -24 y = -4EG = 8 - (-4) = 12

9. JK = JL8m - 5 = 2m + 4 6m = 9 m = 1.5JK = 8(1.5) - 5 = 7

10. WX = WY0.8x + 1.2 = 2.4x 1.2 = 1.6x x = 0.75WY = 2.4(0.75) = 1.8

LESSON 11-2, PAGE 811

11. m � KM = m∠KGL + m∠LGM= m∠KGL + m∠HGJ= 30 + 51 = 81°

12. m � HMK = m∠HGL + m∠LGK= 180 + 30 = 210°

13. m � JK = m∠JGK= m∠HGL - m∠HGJ - m∠KGL= 180 - 51 - 30 = 99°

14. m � MJK = 360 - m � KM = 360 - 81 = 279°

15. Let ST = 2x.x(x) = 4(7 + 11)

x 2 = 72 x = √ �� 72 = 6 √ � 2 ST = 2 (6 √ � 2 ) ≈ 17.0

16. Let CD = 2x. x 2 = 2.5(2.5 + 5)x(x) = 18.75 x = √ �� 18.75 = 2.5 √ � 3 CD = 2 (2.5 √ � 3 ) ≈ 8.7


17. A = π r 2 ( m ____ 360

) = π(12 ) 2 ( 30 ____ 360

) = 12π in. 2 ≈ 37.70 in. 2

18. A = π r 2 ( m ____ 360

) = π(1 ) 2 ( 90 ____ 360

) = 1 _ 4 π m 2 = 0.79 m 2

19. L = 2πr ( m ____ 360

) = 2π(18) ( 160 ____ 360

) = 16π cm ≈ 50.27 cm

20. L = 2πr ( m ____ 360

) = 2π(2) ( 270 ____ 360

) = 3π ft ≈ 9.42 ft


21. m � JL = 2m∠JNL = 2(82) = 164°

22. m∠MKL = 1 _ 2 m � ML = 1 _

2 (64) = 32°

23. ∠B is inscribed in a semicircle is and therefore, a right ∠.3x + 12 = 90 3x = 78 x = 26

24. m∠RSP = 1 _ 2 m � RP = m∠RQP

3y + 3 = 5y - 21 24 = 2y y = 12m∠RSP = 3(12) + 3 = 39°


25. m � MR = 2m∠PMR = 2(41) = 82°

26. m∠QMR = 1 _ 2 m � QR = 1 _

2 (360 - 120 - 82) = 79°

27. m∠GKH = 1 _ 2 (m � FJ + m � GH ) = 1 _

2 (41 + 93) = 67°

28. m∠BXC = 1 _ 2 (m � AD + m � BC )

= 1 _ 2 ( 2 __

16 (360) + 6 __

16 (360)) = 90°


29. BA · AC = DA · AE 3(y) = 7(5) 3y = 35 y = 11 2 _

3

BC = 3 + (11 2 _ 3 ) = 14 2 _

3

DE = 7 + 5 = 12

30. QP · PR = SP · PT z(10) = 15(8) 10z = 120 z = 12QR = (12) + 10 = 22ST = 15 + 8 = 23


31. GJ · HJ = LJ · KJ(4 + 6)6 = (x + 5)5 60 = 5x + 25 35 = 5x x = 7GJ = 4 + 6 = 10LJ = (7) + 5 = 12

32. AB · AC = AD · AE4(4 + y) = 5(5 + 5) 16 + 4y = 50 4y = 34 y = 8 1 _

2

AC = 4 + (8 1 _ 2 ) = 12 1 _

2

AE = 5 + 5 = 10


33. (x - (-4) ) 2 + (y - (-3) ) 2 = 3 2 (x + 4 ) 2 + (y + 3 ) 2 = 9

34. r = √ �� (-2 - (-2) ) 2 + (-2 - 0 ) 2 = 2(x - (-2) ) 2 + (y - 0 ) 2 = 2 2 (x + 2 ) 2 + y 2 = 4

35. (x - 1 ) 2 + (y - (-1) ) 2 = 4 2 (x - 1 ) 2 + (y + 1 ) 2 = 16

36. (x + 2 ) 2 + (y - 2 ) 2 = 1(x - (-2) ) 2 + (y - 2 ) 2 = 1 2 Graph a circle with center (-2, 2) and radius 1.

CHAPTER TEST, PAGE 814

1. chord: −− EC ; tangent: � �� AB ; radii:

−− DE , −− DC ; secant: � �� EC ;

diameter: −− EC

2. Let x be distance to the horizon.(4000 ) 2 + x 2 = (4006.25 ) 2 x 2 = 50,039.0625 x ≈ 224 mi

3. m � JK = m∠JPK = m∠JPL - m∠KPL= m∠MPN - m∠KPL= 84 - 65 = 19°

4. Let UV = 2x.x(x) = 6(9 + 15) x 2 = 144 x = 12 (since x > 0)UV = 2(12) = 24

5. A = π(8 ) 2 ( 135 ____ 360

) = 24π cm 2 ≈ 75.40 cm 2

6. A = 2π(8) ( 135 ____ 360

) = 6π cm ≈ 18.85 cm

7. m � SR = 2m∠SPR = 2(47) = 94° 8. m∠PTQ = 1 _

2 (m � PQ + m � SR ) = 1 _

2 (58 + 94) = 76°

m∠QTR = 180 - m∠PTQ = 180 - 76 = 104° 9. m∠ABC = 180 - 1 _

2 m � AB = 180 - 1 _

2 (128) = 116°

10. m∠MKL = 1 _ 2 (m � JN + m � ML ) = 1 _

2 (118 + 58) = 88°

m∠MKL = 180 - m∠MKL = 180 - 88 = 92°

11. m∠CSD = 1 _ 2 (m � CD - m � AB )

42 = 1 _ 2 (124 - m � AB )

84 = 124 - m � AB m � AB = 124 - 84 = 40°

12. z(2) = 6(4) 2z = 24 z = 12EF = (12) + 2 = 14GH = 4 + 6 = 10

13. 6(6 + x) = 8(8 + 4) 36 + 6x = 96 6x = 60 x = 10PR = 8 + 4 = 12PT = 6 + (10) = 16

14. 4(4) = 2(d - 2) 16 = 2d - 4 20 = 2d d = 10 in.

15. r = √ �� (-2 - 1 ) 2 + (4 - (-2) ) 2 = √ �� 45 = 3 √ � 5

(x - 1 ) 2 + (y - (-2) ) 2 = (3 √ � 5 ) 2

(x - 1 ) 2 + (y + 2 ) 2 = 45

16. Step 1 Plot the 3 given points.Step 2 Connect X, Y, and Z to form a �.

Step 3 Find a point that is equidistant from 3 points by constructing ⊥ bisectors of the 2 sides of �XYZ.The ⊥ bisectors of sides of �XYZ intersect at a point that is equidistant from X, Y, and Z.The intersection of the ⊥ bisectors is P(0, -2). P is center of the circle passing through X, Y, and Z.

COLLEGE ENTRANCE EXAM PRACTICE, PAGE 815

1. EDraw

−− AD . m∠ADC is inscribed in a semicircle. So is a right ∠.m � AB = 2m∠ADB

= 2(90 - m∠BDC) = 2(90 - 30) = 120°

2. APossible coordinates of the center are (1 ± 2, 3 ± 2).(-1, 1) is a possible center. The circle with this center and radius 2 is(x - (-1) ) 2 + (y - 1 ) 2 = 2 2 (x + 1 ) 2 + (y - 1 ) 2 = 4.


3. E PK · KM = LK · KN3(PM - 3) = 6(10 - 6) 3PM - 9 = 24 3PM = 33 PM = 11

4. CL = 2πr ( m � AC _____

360 )

= 2π(6) ( 2m∠ABC ________ 360

)

= 12π ( 50 ____ 360

) = 5 __ 3 π

5. BDraw � with a base along the shaded area and the apex at the center of the circle. � is a 45°-45°-90° � with b = 9 √ � 2 and h = 4.5 √ � 2 .

A = π r 2 ( 90 ____ 360

) - 1 __ 2 bh

= π(9 ) 2 ( 1 __ 4 ) - 1 __

2 (9 √ � 2 ) (4.5 √ � 2 )

= 81 ___ 4 π - 81 ___

2 ≈ 23.12

CHAPTERS 1–11, PAGES 818–819

1. C L = Ph + 1 _

2 P�

328 = 40(4) + 1 _ 2 (40)�

328 = 160 + 20�

168 = 20�

� = 8.4 ft

2. GA = (12 - 2)(3 - 0) - 1 _

2 (12 - 6)(3 - 0)

= 10(3) - 1 _ 2 (6)(3)

= 30 - 9 = 21 units 2

3. Bm∠EFD = m � ED = 45°m∠FED = 180 - (45 + 90) = 45°Since F is the center, �EFC and �CFA are also 45°-45°-90° �.

−− FB ⊥ −− BC . So m � BC = m∠BFC = 45°.

4. G

L = 2πr ( m � ED _____ 360

)

6π = 2πr ( 45 ____ 360

) = 1 __ 4 πr

r = 24 cm

A = π(24 ) 2 ( 45 ____ 360

) = 72π cm 2

5. CF is not on the circle. So

−− AF is not a chord.

6. JKL = 18 and KJ = 24 → tan J = 18 __

24 = 3 _

4 ;

KL 2 + KJ 2 � JL 2 18 2 + 24 2 � 30 2 324 + 576 = 900

7. B m∠P = 1 _

2 (m � RS - m � QU )

= m∠T - 1 _ 2 m � QU

29 = 50 - 1 _ 2 m � QU

1 _ 2 m � QU = 21

m � QU = 42° 8. F

PS · PU = PR · PQ

PS = PR · PQ _______ PU

9. C

From the diagram, y = -3.

10. Gd = √ �� (3 - (-1) ) 2 + (-5 - 1 ) 2 = √ �� 52 = 2 √ �� 13

r = √ �� 13

center = ( -1 + 3 _______ 2 ,

1 + (-5) ________

2 ) = (1, -2)

(x - 1 ) 2 + (y - (-2) ) 2 = ( √ �� 13 ) 2

(x - 1 ) 2 + (y + 2 ) 2 = 13

11. DThe diagonals of a kite intersect at right . So the shortest segment to Q is from the intersection T; −− TQ is shortest segment.

12. HP = 1 _

2 (3b) = 3 ( 1 _

2 b)

A = 1 _ 2 ( 1 _

2 b) ( 1 _

4 b √ � 3 ) = 1 __

16 b 2 √ � 3 = 1 _

4 ( 1 _

2 b ( 1 _

2 b √ � 3 ) )

The area is reduced by 1 _ 4 .

13. CLet the leg length be x m. A = 1 _

2 (x)(x)

36 = 1 _ 2 x 2

72 = x 2

x = 6 √ � 2 hypotenuse = (6 √ � 2 ) √ � 2 = 12 m

14. Let the side lengths be 4x, 5x, and 8x cm. P = 4x + 5x + 8x38.25 = 17x x = 2.254x = 4(2.25) = 9 cm

15. 8 x 2 = 4(16) = 64 x = 8



16. If �HGJ � �LMK, −− HG �

−− LM and −− HJ �

−− LK .

−− HG � −− LM

HG = LM4x + 5 = 13 4x = 8 x = 2Check: HJ = 5(2) - 1 = 9 = KL

17. s = 2, so a = √ � 3 and P = 6(2) = 12

A = 1 _ 2 aP = 1 _

2 ( √ � 3 ) (12) = 6 √ � 3 ≈ 10.4

18. 15.71 mmL = 1 _

2 (2π(5)) = 5π ≈ 15.71 mm

19. 452.39 cm 2 V = 4 _

3 π r 3

288π = 4 _ 3 π r 3

216 = r 3 r = 6 cmS = 4π(6 ) 2 = 144π ≈ 452.39 cm 2

20. 3x = 6 cos 60° = 3

21. Possible answer:180 - 7x = 1 _

2 (4x + 10 + 6x + 14)

180 - 7x = 1 _ 2 (10x + 24)

180 - 7x = 5x + 12 -12x = -168 x = 14

22a. 1st unit:V = 10(5)(9) = 450 ft 3

cost/ ft 3 = 85 ____ 450

≈ $0.19

2nd unit:V = 11(4)(8) = 352 ft 3

cost/ ft 3 = 70 ____ 352

≈ $0.20

The first unit has the lower price per ft 3 .

b. V · $0.25/ ft 3 = $100 V = 400Possible dimensions: 10 ft by 5 ft by 8 ft

23a. x 2 + (y + 1 ) 2 = 25

(x - 0 ) 2 + (y - (-1) ) 2 = 5 2 Plot the circle with center (0, -1) and radius 5.

b. slope of radius to (3, 3) = 3 - (-1)

________ 3 - 0

= 4 __ 3

slope of tangent = - 3 __ 4

equation of tangent: y - 3 = - 3 _

4 (x - 3)

y - 3 = - 3 _ 4 x + 2 1 _

4

y = - 3 _ 4 x + 5 1 _

4

24. The measure of the intercepted arc must be > 0° and < 180°.

25a. Statements Reasons

1. −− AB ‖

−− CD 1. Given2. ∠BCD � ∠ABD 2. Alt. Int. Thm.3. m∠BCD = m∠ABD 3. Def. of � 4. m∠ACD = 1 _

2 m � AD ,

m∠BDC = 1 _ 2 m � BC

4. Inscribed ∠ Thm.

5. 1 _ 2 m � BC = 1 _

2 m � AD 5. Subst. Prop.

6. m � BC = m � AD 6. Mult. Prop. of =

b. Possible answer: Since −− AB ‖

−− CD , m � BC = m � AD from part a. Also, by Inscribed ∠ Thm., m∠ACD

= 1 _ 2 m � AD and m∠BDC = 1 _

2 m � BC . Then use Arc

Add. Post. to get m∠ADC = 1 _ 2 (m � AB + m � BC ) and

m∠BCD = 1 _ 2 (m � AB + m � AD ). Subst. m � BC for m � AD ,

which gives m∠BCD = 1 _ 2 (m � AB + m � BC ) = m∠ADC.

Since ABCD has 1 pair of base , ABCD is isosceles.

c. Possible answer: Since −− AB ‖

−− CD and ABCD is not a trapezoid,

−− AD ‖ −− BC . So ABCD must be a

quadrilateral. Therefore, ∠ADC � ∠ABC. So by def. of �, m∠ADC = m∠ABC. Also, since ABCD can be inscribed in a circle, ∠ADC and ∠ABC are supplementary. So m∠ADC + m∠ABC = 180°. Subst. and solve for m∠ABC: m∠ABC + m∠ABC = 180 2m∠ABC = 180 m∠ABC = 90°Therefore, ∠ABC is a right ∠. Since ABCD is a quadrilateral, it follows that every ∠ is a rt. ∠, so ABCD is a rectangle Therefore, if ABCD is not a trapezoid, it must be a rectangle.


chapter solutions key 11 circles - shakopee.k12.mn.us · by thm. 11-1-3, rs = rt __x 4 = nx-6.3 x =...

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